State Board Primary (1-5) Private Session 2025-26
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KC Sinha: Exercise 4.1 - Mathematics Solution Class 10 Chapter 4 Real numbers
[mathjax] Question 1 From the given figure, find the value of the following: (i) sin C (ii) sin A (iii) cos C (iv) cos A (v) tan C (vi) tan A Sol : (i) Sin C We know that, $\sin \theta=\frac{\text {...
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Question 1
From the given figure, find the value of the following:
(i) sin C (ii) sin A (iii) cos C (iv) cos A (v) tan C (vi) tan A Sol : (i) Sin C We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ So, here θ = C Side opposite to ∠C = AB = 3 Hypotenuse = AC = 5 So, $\sin C=\frac{A B}{A C}=\frac{3}{5}$
(ii) Sin A So, here θ = A The side opposite to ∠A = BC = 4 Hypotenuse = AC = 5 So, $\sin A=\frac{B C}{A C}=\frac{4}{5}$
(iii) Cos C We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ So, here θ = C Side adjacent to ∠C = BC = 4 Hypotenuse = AC = 5 So, $\cos C=\frac{B C}{A C}=\frac{4}{5}$
(iv) Cos A Here, θ = A Side adjacent to ∠A = AB = 3 Hypotenuse = AC = 5 So, $\cos A=\frac{A B}{A C}=\frac{3}{5}$
(v) tan C We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ So, here θ = C Side opposite to ∠C = AB = 3 Side adjacent to ∠C = BC = 4 So, $\tan C=\frac{A B}{B C}=\frac{3}{4}$
(vi) tan A here θ = A Side opposite to ∠A = BC = 4 Side adjacent to ∠A = AB = 3 So, $\tan A=\frac{A B}{B C}=\frac{4}{3}$
Question 2
From the given figure, find the value of :
(i) tan θ (ii) cos θ
Sol : (i) tan θ We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Side opposite to θ = AB = 4 Side adjacent to θ = BC = 3 So, $\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{4}{3}$
(ii) cos θ We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to θ = BC = 3 Hypotenuse = AC = 5 So, $\cos \theta=\frac{B C}{A C}=\frac{3}{5}$
Question 3
From the given figure, find the value of
(i) sin θ (ii) tan θ (iii) tan A – cot C
Sol : (i) sin θ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to θ = BC = ? Hypotenuse = AC = 13 Firstly we have to find the value of BC. So, we can find the value of BC with the help of Pythagoras theorem. According to Pythagoras theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (12)2 + (BC)2 = (13)2 ⇒ 144 + (BC)2 = 169 ⇒ (BC)2 = 169–144 ⇒ (BC)2 = 25 ⇒ BC =√25 ⇒ BC =±5 But side BC can’t be negative. So, BC = 5 Now, BC = 5 and AC = 13 So, $\sin \theta=\frac{B C}{A C}=\frac{5}{13}$
(ii) tan θ We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Side opposite to θ = BC = 5 Side adjacent to θ = AB = 12 So, $\tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
(iii) tan A – cot C We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ and $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ tan A Here, θ = A Side opposite to ∠A = BC = 5 Side adjacent to ∠A = AB = 12 So, $\tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$ Cot C Here, θ = C Side adjacent to ∠C = BC = 5 Side opposite to ∠C = AB = 12 So, $\cot C=\frac{B C}{A B}=\frac{5}{12}$ So, $\tan A-\cot C=\frac{5}{12}-\frac{5}{12}=0$
Question 4 A
In ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
a. sin A, cos A b. sin C, cos C Sol :
(i) (a) sin A We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ So, here θ = A Side opposite to ∠A = BC = 7 Hypotenuse = AC = ? Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem.
According to Pythagoras theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (24)2 + (7)2 = (AC)2 ⇒ 576 + 49 = (AC)2 ⇒ (AC)2 = 625 ⇒ AC =√625 ⇒ AC =±25 But side AC can’t be negative. So, AC = 25cm Now, BC = 7 and AC = 25 So, $\sin A=\frac{B C}{A C}=\frac{7}{25}$ Cos A We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ So, here θ = A Side adjacent to ∠A = AB = 24 Hypotenuse = AC = 25 So,$\cos A=\frac{A B}{A C}=\frac{24}{25}$ (b) sin C
We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ So, here θ = C The side opposite to ∠C = AB = 24 Hypotenuse = AC = 25 So, $\sin C=\frac{A B}{A C}=\frac{24}{25}$ Cos C We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ So, here θ = C Side adjacent to ∠C = BC = 7 Hypotenuse = AC = 25 So, $\cos C=\frac{B C}{A C}=\frac{7}{25}$
Question 4 B
Consider ∆ACB, right angled at C, in which AB = 29 units, BC = 21 units and ∠ABC=θ. Determine the values ofa. cos2 θ+ sin2 θ b. cos2 θ – sin2 θ Sol :(a) Cos2θ +sin2 θ Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem. According to Pythagoras theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ (AC)2 + (BC)2 = (AB)2 ⇒ (AC)2 + (21)2 = (29)2 ⇒ (AC)2 = (29)2 – (21)2 Using the identity a2 –b2 = (a+b) (a – b) ⇒ (AC)2 = (29–21)(29+21) ⇒ (AC)2 = (8)(50) ⇒ (AC)2 = 400 ⇒ AC =√400 ⇒ AC =±20 But side AC can’t be negative. So, AC = 20units Now, we will find the sin θ and cos θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ In ∆ACB, Side opposite to angle θ = AC = 20 and Hypotenuse = AB = 29 So, $\sin \theta=\frac{A C}{A B}=\frac{20}{29}$ Now, We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
In ∆ACB, Side adjacent to angle θ = BC = 21 and Hypotenuse = AB = 29 So, $\cos \theta=\frac{B C}{A B}=\frac{21}{29}$ So$\cos ^{2} \theta+\sin ^{2} \theta=\left(\frac{21}{29}\right)^{2}+\left(\frac{20}{29}\right)^{2}$ $=\frac{441+400}{29 \times 29}$ $=\frac{841}{841}$ =1 Cos2θ +sin2 θ = 1 (b) Cos2θ – sin2 θ Putting values, we get $\cos ^{2} \theta-\sin ^{2} \theta=\left(\frac{21}{29}\right)^{2}-\left(\frac{20}{29}\right)^{2}$ $=\frac{441-400}{29 \times 29}$ $=\frac{41}{841}$
Question 4 C
In ∆ABC, ∠A is a right angle, then find the values of sin B, cos C and tan B in each of the following :a. AB = 12, AC = 5, BC = 13 b. AB = 20, AC = 21, BC = 29 c. BC = √2, AB = AC = 1 Sol : Given that ∠A is a right angle.(a) AB = 12, AC = 5, BC = 13 To Find : sin B, cos C and tan B We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Here, θ = B Side opposite to angle B = AC = 5 Hypotenuse = BC =13 So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{13}$ Now, Cos C We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Here, θ = C Side adjacent to angle C = AC = 5 Hypotenuse = BC =13 So, $\cos C=\frac{A C}{B C}=\frac{5}{13}$ Now, tan B We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Here, θ = B The side opposite to angle B = AC = 5 The side adjacent to angle B = AB = 12 So, $\tan \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{5}{12}$ (b) AB = 20, AC = 21, BC = 29 To Find: sin B, cos C and tan B We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Here, θ = B The side opposite to angle B = AC =21 Hypotenuse = BC =29 So, $\sin B=\frac{A C}{B C}=\frac{21}{29}$ Now, Cos C We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Here, θ = C Side adjacent to angle C = AC = 21 Hypotenuse = BC = 29 So,$\cos C=\frac{A C}{B C}=\frac{21}{29}$
Now, tan B We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Here, θ = B The side opposite to angle B = AC = 21 The side adjacent to angle B = AB = 20 So, $\tan \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{21}{20}$
(c) BC =√2, AB = AC = 1 To Find: sin B, cos C and tan B We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Here, θ = B The side opposite to angle B = AC =1 Hypotenuse = BC =√2 So$\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{1}{\sqrt{2}}$ Now, Cos C We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Here, θ = C Side adjacent to angle C = AC = 1 Hypotenuse = BC = √2 So, $\cos C=\frac{A C}{B C}=\frac{1}{\sqrt{2}}$ Now, tan B We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Here, θ = B The side opposite to angle B = AC = 1 The side adjacent to angle B = AB = 1 So,$\tan B=\frac{A C}{A B}=\frac{1}{1}=1$
Question 5 A
Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :
Sol : Firstly, we give the name to the midpoint of BC i.e. M BC = BM + MC = 2BM or 2MC ⇒ BM = 5 and MC = 5 Now, we have to find the value of AM, and we can find out with the help of Pythagoras theorem. So, In ∆AMB ⇒ (AM)2 + (BM)2 = (AB)2 ⇒ (AM)2 + (5)2 = (13)2 ⇒ (AM)2 = (13)2 – (5)2 Using the identity a2 –b2 = (a+b) (a – b) ⇒ (AM)2 = (13–5)(13+5) ⇒ (AM)2 = (8)(18) ⇒ (AM)2 = 144 ⇒ AM =√144 ⇒ AM =±12 But side AM can’t be negative. So, AM = 12 a. sin θ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ In ∆AMB Side opposite to θ = AM = 12 Hypotenuse = AB=13 So, $\sin \theta=\frac{A M}{A B}=\frac{12}{13}$ So, $\sin \theta=\frac{12}{13}$ b. cos θ We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ In ∆AMB The side adjacent to θ = BM = 5 Hypotenuse = AB = 13 So, $\cos \theta=\frac{B M}{A B}=\frac{5}{13}$ So, $\cos \theta=\frac{5}{13}$ c. tan θ We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ In ∆AMB Side opposite to θ = AM = 12 The side adjacent to θ = BM = 5 So, $\tan \theta=\frac{\mathrm{AM}}{\mathrm{BM}}=\frac{12}{5}$ So, $\tan \theta=\frac{12}{5}$
Question 5 B
Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below : Sol : Firstly, we have to find the value of XM and we can find out with the help of Pythagoras theorem So, In ∆XMZ ⇒ (XM)2 + (MZ)2 = (XZ)2 ⇒ (XM)2 + (16)2 = (20)2 ⇒ (XM)2 = (20)2 – (16)2 Using the identity a2 –b2 = (a+b) (a – b) ⇒ (XM)2 = (20–16)(20+16) ⇒ (XM)2 = (4)(36) ⇒ (XM)2 = 144 ⇒ XM =√144 ⇒ XM =±12 But side XM can’t be negative. So, XM = 12 Now, In ∆XMY we have the value of XM and MY but we don’t have the value of XY. So, again we apply the Pythagoras theorem in ∆XMY ⇒ (XM)2 + (MY)2 = (XY)2 ⇒ (12)2 + (5)2 = (XY)2 ⇒ 144 + 25 = (XY)2 ⇒ (XY)2 = 169 ⇒ XY =√169 ⇒ XY =±13 But side XY can’t be negative. So, XY = 13 a. sin θ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ In ∆XMY Side opposite to θ = MY = 5 Hypotenuse = XY = 13 So, $\sin \theta=\frac{\mathrm{MY}}{\mathrm{XY}}=\frac{5}{13}$ b. cos θ We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ In ∆XMY Side adjacent to θ = XM = 12 Hypotenuse = XY = 13 So,$\cos \theta=\frac{\mathrm{XM}}{\mathrm{XY}}=\frac{12}{13}$ c. tan θ We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ In ∆XMY The side opposite to θ = MY = 5 Side adjacent to θ = XM = 12 So, $\tan \theta=\frac{\mathrm{MY}}{\mathrm{XM}}=\frac{5}{12}$
Question 6
In ∆PQR, ∠Q is a right angle PQ = 3, QR = 4. If ∠P=α and ∠R=β, then find the values of(i) sin α (ii) cos α (iii) tan α (iv) sin β (v) cos β (vi) tan β Sol : Given : PQ = 3, QR = 4 ⇒ (PQ)2 + (QR)2 = (PR)2 ⇒ (3)2 + (4)2 = (PR)2 ⇒ 9 + 16 = (PR)2 ⇒ (PR)2 = 25 ⇒ PR =√25 ⇒ PR =±5 But side PR can’t be negative. So, PR = 5 (i) sin α We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Here, θ = α The side opposite to angle α = QR =4 Hypotenuse = PR =5 So, $\sin \alpha=\frac{4}{5}$ (ii) cos α We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Here, θ = α The side adjacent to angle α = PQ =3 Hypotenuse = PR =5 So, $\cos \alpha=\frac{3}{5}$ (iii) tan α We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Here, θ = α Side opposite to angle α = QR =4 Side adjacent to angle α = PQ =3 So,$\tan \alpha=\frac{4}{3}$ (iv) sin β We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Here, θ = β The side opposite to angle β = PQ =3 Hypotenuse = PR =5 So, $\sin \beta=\frac{3}{5}$ (v) cos β We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Here, θ = β Side adjacent to angle β = QR =4 Hypotenuse = PR =5 So, $\cos \beta=\frac{4}{5}$ (vi) tan β We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Here, θ = β Side opposite to angle β = PQ =3 Side adjacent to angle β = QR =4 So, $\tan \beta=\frac{3}{4}$
Question 7 A
If $\sin \theta=\frac{4}{5}$ ,then find the values of cos θ and tan θ.Sol : Given: $\sin \theta=\frac{4}{5}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \theta=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$ Let, Perpendicular =AB =4k and Hypotenuse =AC =5k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle In right angled ∆ ABC, we have ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (4k)2 + (BC)2 = (5k)2 ⇒ 16k2 + (BC)2 = 25k2 ⇒ (BC)2 = 25 k2 –16 k2 ⇒ (BC)2 = 9 k2 ⇒ BC =√9 k2 ⇒ BC =±3k But side BC can’t be negative. So, BC = 3k Now, we have to find the value of cos θ and tan θ We know that, $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$ The side adjacent to angle θ or base = BC =3k Hypotenuse = AC =5k So, $\cos \theta=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$ Now, We know that,$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Perpendicular = AB =4k Base = BC =3k So, $\tan \theta=\frac{4 k}{3 k}=\frac{4}{3}$
Question 7 B
If $\sin \mathrm{A}=\frac{3}{4}$ ,calculate cos A and tan A.Sol : Given: Sin A $=\frac{3}{4}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{4} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}$ Let, Side opposite to angle θ = BC =3k and Hypotenuse = AC =4k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (AB)2 + (3k)2 = (4k)2 ⇒ (AB)2 + 9k2 = 16k2 ⇒ (AB)2 = 16 k2 – 9 k2 ⇒ (AB)2 = 7 k2 ⇒ AB =k√7 So, AB = k√7 Now, we have to find the value of cos A and tan A We know that, $\cos \theta=\frac{\text { Side adjacent to angle } \theta}{\text { hypotenuse }}$ Here, θ = A The side adjacent to angle A = AB =k√7 Hypotenuse = AC =4k So,$\cos A=\frac{k \sqrt{7}}{4 k}=\frac{\sqrt{7}}{4}$ Now, We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ The side opposite to angle A = BC =3k The side adjacent to angle A = AB =k√7 So,$\tan \mathrm{A}=\frac{3 \mathrm{k}}{\mathrm{k} \sqrt{7}}=\frac{3}{\sqrt{7}}$
Question 8
If $\sin \theta=\frac{3}{5}$ , then find the values cos θ and tan θ.Sol : Given:$\sin \theta=\frac{3}{5}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$ Let, Perpendicular =AB =3k and Hypotenuse =AC =5k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (BC)2 = (5k)2 ⇒ 9k2 + (BC)2 = 25k2 ⇒ (BC)2 = 25 k2 – 9 k2 ⇒ (BC)2 = 16 k2 ⇒ BC =√16 k2 ⇒ BC =±4k But side BC can’t be negative. So, BC = 4k Now, we have to find the value of cos θ and tan θ We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle θ = BC =4k Hypotenuse = AC =5k So, $\cos \theta=\frac{4 \mathrm{k}}{5 \mathrm{k}}=\frac{4}{5}$ Now, tan θ We know that, $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Perpendicular = AB =3k Base = BC =4k So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$
Question 9
If $\cos \theta=\frac{4}{5}$ , then find the value of tan θ.Sol : We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$ $\cos \theta=\frac{4}{5} \Rightarrow \frac{B}{H}=\frac{4}{5} \Rightarrow \frac{B C}{A C}=\frac{4}{5}$ Let, Base =BC = 4k Hypotenuse =AC = 5k Where, k ia any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (AB)2 + (4k)2 = (5k)2 ⇒ (AB)2 + 16k2 = 25k2 ⇒ (AB)2 = 25 k2 –16 k2 ⇒ (AB)2 = 9 k2 ⇒ AB =√9 k2 ⇒ AB =±3k But side AB can’t be negative. So, AB = 3k Now, we have to find tan θ We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Side opposite to angle θ = BC =4k Side adjacent to angle θ = AB =3k So,$\tan \theta=\frac{4 k}{3 k}=\frac{4}{3}$
Question 10 A
If $\tan \theta=\frac{3}{4}$ then find the values of cos θ and sin θ.Sol :We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ $\tan \theta=\frac{3}{4}$$\Rightarrow \frac{P}{B}=\frac{3}{4}$$\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$ Let, The side opposite to angle θ =AB = 3k The side adjacent to angle θ =BC = 4k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (4k)2 = (AC)2 ⇒ (AC)2 = 9 k2+16 k2 ⇒ (AC)2 = 25 k2 ⇒ AC =√25 k2 ⇒ AC =±5k But side AC can’t be negative. So, AC = 5k Now, we will find the sin θ and cos θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = AB = 3k and Hypotenuse = AC = 5k So, $\sin \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$ Now, We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = BC = 4k and Hypotenuse = AC = 5k So,$\cos \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
Question 10 B
If tan A= 4/3. Find the other trigonometric ratios of the angle A.Sol :
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Here, θ = A$\tan \mathrm{A}=\frac{4}{3}$$\Rightarrow \frac{P}{B}=\frac{4}{3}$$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}$Let, The side opposite to angle A =BC = 4k The side adjacent to angle A =AB = 3k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (4k)2 = (AC)2 ⇒ (AC)2 = 9 k2 +16 k2 ⇒ (AC)2 = 25 k2 ⇒ AC =√25 k2 ⇒ AC =±5k But side AC can’t be negative. So, AC = 5k Now, we will find the sin A and cos A $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle A = BC = 4k and Hypotenuse = AC = 5k So, $\sin A=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
Now, We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle A = AB = 3k and Hypotenuse = AC = 5k
So, $\cos A=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$ Now, we find other trigonometric ratios $\operatorname{cosec} \mathrm{A}=\frac{1}{\sin \mathrm{A}}$ $=\frac{1}{\frac{4}{5}}$ $=\frac{5}{4}$$\sec \mathrm{A}=\frac{1}{\cos \mathrm{A}}$ $=\frac{1}{\frac{3}{5}}$ $=\frac{5}{3}$$\cot A=\frac{1}{\tan A}$ $=\frac{1}{\frac{4}{3}}$ $=\frac{3}{4}$
Question 11
If cot $\theta=\frac{12}{5}$, then find the value of sin θ.Sol : We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \theta=\frac{12}{5} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{12}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12}{5}$ Let, Side adjacent to angle θ =AB = 12k The side opposite to angle θ =BC = 5k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (12k)2 + (5k)2 = (AC)2 ⇒ (AC)2 = 144 k2 +25 k2 ⇒ (AC)2 = 169 k2 ⇒ AC =√169 k2 ⇒ AC =±13k But side AC can’t be negative. So, AC = 13k Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = BC = 5k and Hypotenuse = AC = 13k So, $\sin \theta=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$
Question 12
If tan $\theta=\frac{5}{12}$, then find the value of cos θ.Sol : We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ $\tan \theta=\frac{5}{12} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{5}{12} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$ Let, The side opposite to angle θ =BC = 5k The side adjacent to angle θ =AB = 12k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (12k)2 + (5k)2 = (AC)2 ⇒ (AC)2 = 144 k2 +25 k2 ⇒ (AC)2 = 169 k2 ⇒ AC =√169 k2 ⇒ AC =±13k But side AC can’t be negative. So, AC = 13k Now, We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = AB = 12k and Hypotenuse = AC = 13k So, $\cos \theta=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$
Question 13
If sin $\theta=\frac{12}{13}$, then find the value of cos θ and tan θ.Sol : Given: $\sin \theta=\frac{12}{13}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \theta=\frac{12}{13} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{12}{13} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}$ Let, Side opposite to angle θ = 12k and Hypotenuse = 13k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (12k)2 + (BCk)2 = (13)2 ⇒ 144 k2 + (BC)2 = 169 k2 ⇒ (BC)2 = 169 k2 –144 k2 ⇒ (BC)2 = 25 k2 ⇒ BC =√25 k2 ⇒ BC =±5k But side BC can’t be negative. So, BC = 5k Now, we have to find the value of cos θ and tan θ We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = BC =5k Hypotenuse = AC =13k So, $\cos \theta=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$ Now, tan θ We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ side opposite to angle θ = AB =12k Side adjacent to angle θ = BC =5k So, $\tan \theta=\frac{12 \mathrm{k}}{5 \mathrm{k}}=\frac{12}{5}$
Question 14
If tan θ =0.75, then find the value of sin θ.Sol :We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$Given: tan θ =0.75 $\Rightarrow \tan \theta=\frac{75}{100}=\frac{3}{4}$ $\tan \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{3}{4} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$ Let, The side opposite to angle θ =BC = 3k The side adjacent to angle θ =AB = 4k Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (4k)2 + (3k)2 = (AC)2 ⇒ (AC)2 = 16 k2 +9 k2 ⇒ (AC)2 = 25 k2 ⇒ AC =√25 k2 ⇒ AC =±5k But side AC can’t be negative. So, AC = 5k Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = BC = 3k and Hypotenuse = AC = 5k So, $\sin \theta=\frac{B C}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Question 15
If tan B= √3, then find the values of sin B and cos B.Sol :We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Given: tan B = √3 $\Rightarrow \tan \mathrm{B}=\frac{\sqrt{3}}{1}$ $\tan \mathrm{B}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$ Let, Side opposite to angle B =AC = √3k The side adjacent to angle B =AB = 1k where k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (AC)2 = (BC)2 ⇒ (1k)2 + (√3k)2 = (BC)2 ⇒ (BC)2 = 1 k2 +3 k2 ⇒ (BC)2 = 4 k2 ⇒ BC =√2 k2 ⇒ BC =±2k But side BC can’t be negative. So, BC = 2k Now, we will find the sin B and cos B $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle B = AC = k√3 and Hypotenuse = BC = 2k So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$ Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle B = AB =1k Hypotenuse = BC =2k So, $\cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$
Question 16
If $\tan \theta=\frac{\mathrm{m}}{\mathrm{n}}$, then find the values of cos θ and sin θ.Sol :We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Here, $\tan \theta=\frac{\mathrm{m}}{\mathrm{n}}$ So, Side opposite to angle θ =AC = m The side adjacent to angle θ =AB = n Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (AC)2 = (BC)2 ⇒ (n)2 + (m)2 = (BC)2 ⇒ (BC)2 = m2 + n2 ⇒ BC =√ m2 + n2 So, BC =√(m2 + n2) Now, we will find the sin B and cos B $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = AC = m and Hypotenuse = BC =√(m2 + n2) So, $\sin \theta=\frac{A C}{B C}=\frac{m}{\sqrt{m^{2}+n^{2}}}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = AB =n Hypotenuse = BC =√(m2 + n2) So, $\cos \theta=\frac{A B}{B C}=\frac{n}{\sqrt{m^{2}+n^{2}}}$
Question 17
If sin θ = √3 cos θ, then find the values of cos θ and sin θ.Sol : Given : sin θ =√3cos θ $\Rightarrow \frac{\sin \theta}{\cos \theta}=\sqrt{3}$ ⇒ tan θ =√3 We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ and tan θ = √3 $\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}$ $\tan \theta=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$ Let, The side opposite to angle θ =AC = k√3 The side adjacent to angle θ =AB = 1k where k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (AC)2 = (BC)2 ⇒ (1k)2 + (k√3)2 = (BC)2 ⇒ (BC)2 = 1 k2 +3 k2 ⇒ (BC)2 = 4 k2 ⇒ BC =√2 k2 ⇒ BC =±2k But side BC can’t be negative. So, BC = 2k Now, we will find the sin θ and cos θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = AC = k√3 and Hypotenuse = BC = 2k So,$\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$ Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle θ = AB =1k Hypotenuse = BC =2k So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$
Question 18 A
If $\cot \theta=\frac{21}{20}$, then find the values of cos θ and sin θ.Sol : We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \theta=\frac{21}{20} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{21}{20} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{21}{20}$ Let, Side adjacent to angle θ =AB = 21k The side opposite to angle θ =BC = 20k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (21k)2 + (20k)2 = (AC)2 ⇒ (AC)2 = 441 k2 +400 k2 ⇒ (AC)2 = 841 k2 ⇒ AC =√841 k2 ⇒ AC =±29k But side AC can’t be negative. So, AC = 29k Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = BC = 20k and Hypotenuse = AC = 29k So, $\sin \theta=\frac{B C}{A C}=\frac{20 k}{29 k}=\frac{20}{29}$ Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = AB =21k Hypotenuse = AC =29k So, $\cos \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{21 \mathrm{k}}{29 \mathrm{k}}=\frac{21}{29}$
Question 18 B
If 15 cot A=18, find sin A and sec A.Sol : Given: 15 cot A = 8 $\Rightarrow \cot A=\frac{8}{15}$ And we know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \mathrm{A}=\frac{8}{15} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{8}{15} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8}{15}$ Let, Side adjacent to angle A =AB = 8k The side opposite to angle A =BC = 15k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (8k)2 + (15k)2 = (AC)2 ⇒ (AC)2 = 64 k2 +225 k2 ⇒ (AC)2 = 289 k2 ⇒ AC =√289 k2 ⇒ AC =±17k But side AC can’t be negative. So, AC = 17k Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = BC = 15k and Hypotenuse = AC = 17k So, $\sin \theta=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$ Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle θ = AB =8 Hypotenuse = AC =17 So, $\cos \theta=\frac{A B}{A C}=\frac{8 k}{17 k}=\frac{8}{17}$ $\therefore \sec \theta=\frac{1}{\cos \theta}$$=\frac{1}{\frac{8}{17}}$$=\frac{17}{8}$
Question 19
If sin θ = cos θ and 0° < θ <90°, then find the values of sin θ and cos θ.Sol : Given: sinθ = cosθ $\Rightarrow \frac{\sin \theta}{\cos \theta}=1$ ⇒ tan θ = 1$\tan \theta=\frac{1}{1} \Rightarrow \frac{P}{B}=\frac{1}{1} \Rightarrow \frac{A B}{B C}=\frac{1}{1}$ Let, Side opposite to angle θ = AB =1k The side adjacent to angle θ = BC =1k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (1k)2 + (1k)2 = (AC)2 ⇒ (AC)2 = 1k2 +1k2 ⇒ (AC)2 = 2k2 ⇒ AC =√2k2 ⇒ AC =k√2 So, AC = k√2 Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = AB= 1k and Hypotenuse = AC = k√2 So, $\sin \theta=\frac{A B}{A C}=\frac{1 k}{k \sqrt{2}}=\frac{1}{\sqrt{2}}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle θ = BC =1k Hypotenuse = AC =k√2 So, $\cos \theta=\frac{B C}{A C}=\frac{1 k}{k \sqrt{2}}=\frac{1}{\sqrt{2}}$
Question 20
If $\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$, then find the values of cos θ and $\frac{1}{\tan \theta}$.Sol : $\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+v^{2}}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or, $\sin \theta=\frac{\text { Perpendicular }}{\text { hypotenuse }}$$\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ $\Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}}$ $ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}}$Let, Side opposite to angle θ = AB = x2 – y2 and Hypotenuse = AC = x2 + y2 In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (x2 – y2 )2 + (BC)2 = (x2 + y2 )2 ⇒ (BC)2 = (x2 + y2 )2 – (x2 – y2 )2 Using the identity, a2 – b2 = (a+b)(a – b) ⇒ (BC)2 = [(x2 + y2 + x2 – y2 )][ x2 + y2 –( x2 – y2)] ⇒ (BC)2 = (2x2)(2y2) ⇒ (BC)2 = (4x2y2) ⇒ BC =√4x2y2 ⇒ BC = ±2xy ⇒ BC = 2xy [taking positive square root since, side cannot be negative] $\therefore \cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{\text { BC }}{\text { AC }}=\frac{2 x y}{x^{2}+y^{2}}$ and $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}$ So, $\frac{1}{\tan \theta}=\frac{1}{\frac{x^{2}-y^{2}}{2 x y}}=\frac{2 x y}{x^{2}-y^{2}}$
Question 21
Iftan $\theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$, then find the values of sin θ and cos θ.Sol : Given: $\tan \theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$ We know that, $\tan \theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$ Let, AB = √(m2 – n2) and BC = n In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (√(m2 – n2))2 + (n)2 = (AC )2 ⇒ m2 – n2 + n2 = (AC )2 ⇒ (AC)2 = (m2) ⇒ AC =√ m2 ⇒ AC = ±m ⇒ AC = m [taking positive square root since, side cannot be negative] Now, we have to find the value of cos θ and sin θ We, know that $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{n}}{\mathrm{m}}$ and $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $=\frac{A B}{A C}=\frac{\sqrt{m^{2}-n^{2}}}{m}$
Question 22 A
If sec θ = 2, then find the values of other t–ratios of angle θ.Sol : Given: sec θ = 2We know that, $\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$ $\operatorname{Sec} \theta=\frac{2}{1} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{2}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{1}$ Let, BC = 1k and AC = 2k where, k is any positive integer. In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (AB)2 + (1k)2 = (2k )2 ⇒ (AB)2 + k2 = 4k2 ⇒ (AB)2 = 4k2 – k2 ⇒ (AB)2 = 3k2 ⇒ AB = k√3 Now, we have to find the value of other trigonometric ratios. We, know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $=\frac{A B}{A C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$ $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$ $=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{1 \mathrm{k}}=\frac{\sqrt{3}}{1}=\sqrt{3}$ $\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$ $\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\sqrt{3}}$
Question 22 B
Given $\sec \theta=\frac{13}{12}$ calculate all other trigonometric ratios.Sol :
Given: $\sec \theta=\frac{13}{12}$
We know that, $\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$ $\sec \theta=\frac{13}{12} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{13}{12} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{12}$ Let, BC = 12k and AC = 13k where, k is any positive integer. In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (AB)2 + (12k)2 = (13k )2 ⇒ (AB)2 + 144k2 = 169k2 ⇒ (AB)2 = 169k2 – 144k2 ⇒ (AB)2 = 25k2 ⇒ AB = √25k2 ⇒ AB =±5k [taking positive square root since, side cannot be negative] Now, we have to find the value of other trigonometric ratios. We, know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $=\frac{A B}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$ $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $=\frac{B C}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$ $=\frac{A B}{B C}=\frac{5 k}{12 k}=\frac{5}{12}$$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{5}{13}}=\frac{13}{5}$$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{5}{12}}=\frac{12}{5}$
Question 23
If $\operatorname{cosec} \theta=\sqrt{10}$, then find the values of other t–ratios of angle θ.Sol : We know that, $\operatorname{cosec} \theta=\frac{\text { hypotenuse }}{\text { perpendicular }}$ $\operatorname{cosec} \theta=\frac{\sqrt{10}}{1} \Rightarrow \frac{\mathrm{H}}{\mathrm{P}}=\frac{\sqrt{10}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{10}}{1}$ Let, AB = 1k and AC = k√10 where, k is any positive integer. In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (1k )2+ (BC)2 = (k√10)2 ⇒ (BC)2 = 10k2 – k2 ⇒ (BC)2 = 9k2 ⇒ BC = √9k2 ⇒ BC =±3k [taking positive square root since, side cannot be negative] Now, we have to find the value of other trigonometric ratios. We, know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{\mathrm{k} \sqrt{10}}=\frac{1}{\sqrt{10}}$$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $=\frac{B C}{A C}=\frac{3 k}{k \sqrt{10}}=\frac{3}{\sqrt{10}}$$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$ $=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1 \mathrm{k}}{3 \mathrm{k}}=\frac{1}{3}$$\sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{\sqrt{10}}}=\frac{\sqrt{10}}{3}$$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{1}{3}}=3$
Question 24 A
If $\tan \mathrm{A}=\frac{\sqrt{3}}{2}$ then find the values of sin A + cos A.Sol :We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Given: $\tan \mathrm{A}=\frac{\sqrt{3}}{2}$ $\Rightarrow \tan \mathrm{A}=\frac{\sqrt{3}}{2}$ $\tan \mathrm{A}=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\sqrt{3}}{2}$ Let, Side opposite to angle A =BC = k√3 Side adjacent to angle A =AB = 2k where, k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (2k)2 + (√3k)2 = (AC)2 ⇒ (AC)2 = 4 k2 +3 k2 ⇒ (AC)2 = 7 k2 ⇒ AC =√7 k2 ⇒ AC =k√7 So, AC = k√7
Now, we will find the sin A and cos A $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle A = BC = k√3 and Hypotenuse = AC = k√7 So, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{k} \sqrt{3}}{\mathrm{k} \sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle A = AB =2k Hypotenuse = AC = k√7 So, $\cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2 \mathrm{k}}{\mathrm{k} \sqrt{7}}=\frac{2}{\sqrt{7}}$ Now, we have to find sin A +cos A
Putting values of sin A and cos A, we get $\sin \mathrm{A}+\cos \mathrm{A}=\frac{\sqrt{3}}{\sqrt{7}}+\frac{2}{\sqrt{7}}=\frac{\sqrt{3}+2}{\sqrt{7}}$
Question 24 B
If $\sin \theta=\sqrt{3} \cos \theta$ find the value of cos θ – sin θ.Sol : Given: sin θ =√3cos θ $\Rightarrow \frac{\sin \theta}{\cos \theta}=\sqrt{3}$ ⇒ tan θ = √3We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Given: tan θ = √3 $\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}$ $\tan \theta=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$ Let, Side opposite to angle θ =AC = √3k Side adjacent to angle θ =AB = 1k where k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (AC)2 = (BC)2 ⇒ (1k)2 + (√3k)2 = (BC)2 ⇒ (BC)2 = 1 k2 +3 k2 ⇒ (BC)2 = 4 k2 ⇒ BC =√2 k2 ⇒ BC =±2k But side BC can’t be negative. So, BC = 2k
Now, we will find the sin B and cos B $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = AC = k√3 and Hypotenuse = BC = 2k So, $\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle θ = AB =1k Hypotenuse = BC =2k So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$ Now, we have to find the value of cos θ – sin θ Putting the values of sin θ and cos θ, we get $\cos \theta-\sin \theta=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}$
Question 24 C
If $\tan \theta=\frac{8}{15}$, find the value of 1+ cos2 θ.Sol :We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ $\tan \theta=\frac{8}{15} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{8}{15} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{8}{15}$ Let, Side opposite to angle θ =AB = 8k Side adjacent to angle θ =BC = 15k where, k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (8k)2 + (15k)2 = (AC)2 ⇒ (AC)2 = 64k2+225k2 ⇒ (AC)2 = 289 k2 ⇒ AC =√289 k2 ⇒ AC =±17k But side AC can’t be negative. So, AC = 17k Now, we will find the cos θ
We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = BC = 15k and Hypotenuse = AC = 17k So, $\cos \theta=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$ Now, we have to find the value of 1+ cos2 θ
Putting the value of cos θ, we get $1+\cos ^{2} \theta=1+\left(\frac{15}{17}\right)^{2}$$=1+\frac{225}{289}$$=\frac{289+225}{289}$$=\frac{514}{289}$
Question 25
If $\cot \theta=\frac{7}{8}$, evaluate(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ (ii) $\cot ^{2} \theta$Sol :Given: $\cot \theta=\frac{7}{8}$ We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \theta=\frac{7}{8} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{7}{8} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{7}{8}$ Let, Side adjacent to angle θ =AB = 7k Side opposite to angle θ =BC = 8k where, k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (7k)2 + (8k)2 = (AC)2 ⇒ (AC)2 = 49 k2 +69 k2 ⇒ (AC)2 = 113 k2 ⇒ AC =√113 k2 ⇒ AC =k√113 $\therefore \sin \theta=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8 \mathrm{k}}{\mathrm{k} \sqrt{113}}=\frac{8}{\sqrt{113}}$and $\cos \theta=\frac{B}{H}=\frac{A B}{A C}=\frac{7 k}{k \sqrt{113}}=\frac{7}{\sqrt{113}}$(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ We know that, (a+b)(a – b) = (a2 – b2) So, using this identity, we get $=\frac{(1)^{2}-(\sin \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$$=\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta}$$=\frac{1-\left(\frac{8}{\sqrt{113}}\right)^{2}}{1-\left(\frac{7}{\sqrt{113}}\right)^{2}}$$=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}$$=\frac{\frac{113-64}{113}}{\frac{113-49}{113}}$$=\frac{49}{64}$(ii) cot2 θ Given $\cot \theta=\frac{7}{8}$$=\left(\frac{7}{8}\right)^{2}$$=\frac{49}{64}$
Question 26 A
If 3 cot A = 4, check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$=cos2A–sin2 A or not.Sol : Given: 3cot A = 4 $\Rightarrow \cot A=\frac{4}{3}$We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot A=\frac{4}{3} \Rightarrow \frac{B}{P}=\frac{4}{3} \Rightarrow \frac{A B}{B C}=\frac{4}{3}$ Let, Side adjacent to angle A =AB = 4k The side opposite to angle A =BC = 3k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (4k)2 + (3k)2 = (AC)2 ⇒ (AC)2 = 16 k2 + 9 k2 ⇒ (AC)2 = 25 k2 ⇒ AC =√25k2 ⇒ AC = ±5k [taking positive square root since, side cannot be negative] $\therefore \tan \mathrm{A}=\frac{1}{\cot \mathrm{A}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$ $\sin \mathrm{A}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$ and $\cos A=\frac{B}{H}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
And RHS = cos2 A – sin2 A $=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$$=\frac{16}{25}-\frac{9}{25}$$=\frac{7}{25}$ …(ii)
From Eqs. (i) and (ii) LHS =RHS Hence Proved
Question 26 B
In a right triangle ABC, right angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.Sol : tan A = 1 As we know $\tan \theta=\frac{\text { perpedicular }}{\text { base }}$ Now construct a right angle triangle right angled at B such that ∠ BAC = θ Hence perpendicular = BC = 1 and base = AB = 1By Pythagoras theorem, AC2 = AB2 + BC2 ⇒ AC2 = (1)2 + (1)2 ⇒ AC2 = 2 ⇒ AC = As,$\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}$ and $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$ ⇒ $\sin \theta=\frac{1}{\sqrt{2}}$ and $\cos \theta=\frac{1}{\sqrt{2}}$ Hence, 2 sin A cos A=$2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$ ⇒ 2 sin A cos A=$2 \times \frac{1}{2}$ ⇒ 2 sin A cos A=1 = R.H.S Hence proved.
Question 27
If 4sin2 θ =3 and 0o < θ <90o, find the value of 1 + cos θ.Sol : 4sin2 θ =3 $\Rightarrow \sin ^{2} \theta=\frac{3}{4}$ $\Rightarrow \sin \theta=\pm \frac{\sqrt{3}}{2}$
But it is given 0o< θ <90o So, $\sin \theta=\frac{\sqrt{3}}{2}$ $\sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\sqrt{3}}{2}$
Let, P =k√3 and H =2k In right angled ∆ABC, we have B2 + P2 = H2 ⇒ B2 + (k√3)2 = (2k)2 ⇒ B2 + 3k2 = 4k2 ⇒ B2 = 4k2 – 3k2 ⇒ B2 = k2 ⇒ B = ±k ⇒ B = k [taking positive square root since, side cannot be negative] $\therefore \cos \theta=\frac{B}{H}=\frac{k}{2 k}=\frac{1}{2}$ So, $1+\cos \theta=1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}$
Question 28
If $\tan \theta=\frac{p}{q}$find the value of $\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}$.Sol : Given:$\tan \theta=\frac{p}{q}$
If 13 cos θ = 5, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$.Sol : Given: 13 cosθ = 5 $\Rightarrow \cos \theta=\frac{5}{13}$We know that, $\cos \theta=\frac{\text { Base }}{\text { hypotenuse }}$ $\cos \theta=\frac{5}{13} \Rightarrow \frac{B}{H}=\frac{5}{13}$Let AB =5k and BC = 13k In right angled ∆ABC, we have B2 + P2 = H2 ⇒ (5k)2 + P2 = (13k)2 ⇒ P2 + 25k2 = 169k2 ⇒ P2 = 169k2 – 25k2 ⇒ P2 = 144k2 ⇒ P =√144k2 ⇒ P = ±12k ⇒ P = 12k [taking positive square root since, side cannot be negative] $\therefore \sin \theta=\frac{P}{H}=\frac{12}{13}$Now, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$ $=\frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}-\frac{5}{13}}$ $=\frac{17}{7}$
Question 30
If $\sec \theta=\frac{13}{5}$, show that $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}=3$.Sol : Given: $\sec \theta=\frac{13}{5}$We know that, $\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$ $\operatorname{Sec} \theta=\frac{13}{5} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{13}{5} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{5}$ Let, BC = 5k and AC = 13k where, k is any positive integer. In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (AB)2 + (5k)2 = (13k )2 ⇒ (AB)2 + 25k2 = 169k2 ⇒ (AB)2 = 169k2 – 25k2 ⇒ (AB)2 = 144k2 ⇒ AB = √144k2 ⇒ AB =±12k [taking positive square root since, side cannot be negative] Now, we have to find the value of other trigonometric ratios. We, know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$
If 2 tan θ = 1, find the value of $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$.Sol : Given: 2 tan θ = 1 $\Rightarrow \tan \theta=\frac{1}{2}$We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ $\tan \theta=\frac{1}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{1}{2} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{2}$ Let, Side opposite to angle θ =AB = 1k Side adjacent to angle θ =BC = 2k where, k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (k)2 + (2k)2 = (AC)2 ⇒ (AC)2 = k2+4k2 ⇒ (AC)2 = 5k2 ⇒ AC =√5k2 ⇒ AC =±k√5 But side AC can’t be negative. So, AC = k√5
Now, we will find the sin θ and cos θ
We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = BC = 2k and Hypotenuse = AC = k√5 So,$\cos \theta=\frac{B C}{A C}=\frac{2 k}{k \sqrt{5}}=\frac{2}{\sqrt{5}}$ And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ =AB = 1k And Hypotenuse =AC = k√5 So, $\sin \theta=\frac{A B}{A C}=\frac{1 k}{k \sqrt{5}}=\frac{1}{\sqrt{5}}$ Now, $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$ $=\frac{3\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}{2\left(\frac{2}{\sqrt{5}}\right)-\frac{1}{\sqrt{5}}}$
$=\frac{6+1}{4-1}$
$=\frac{7}{3}$
Question 32
If 5 tan α = 4, show that $\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}=\frac{1}{6}$.Sol : Given: 5 tan = 4 $\Rightarrow \tan \alpha=\frac{4}{5}$We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ $\tan \alpha=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{4}{5}$Let, The side opposite to angle α =AB = 4k The side adjacent to angle α =BC = 5k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (4k)2 + (5k)2 = (AC)2 ⇒ (AC)2 = 16k2+25k2 ⇒ (AC)2 = 41k2 ⇒ AC =√41k2 ⇒ AC =±k√41 But side AC can’t be negative. So, AC = k√41 Now, we will find the sin α and cos α We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle α = BC = 5k and Hypotenuse = AC = k√41 So, $\cos \alpha=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 \mathrm{k}}{\mathrm{k} \sqrt{41}}=\frac{5}{\sqrt{41}}$ And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle α =AB = 4k And Hypotenuse =AC = k√5 So, $\sin \alpha=\frac{A B}{A C}=\frac{4 k}{k \sqrt{4} 1}=\frac{4}{\sqrt{41}}$
If $\cot \theta=\frac{3}{4}$prove that $\sqrt{\frac{\sec \theta+\operatorname{cosec} \theta}{\sec \theta-\operatorname{cosec} \theta}}=\sqrt{7}$.Sol :We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{3}{4} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$ Let, Side adjacent to angle θ =AB = 3k The side opposite to angle θ =BC = 4k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (4k)2 = (AC)2 ⇒ (AC)2 = 9k2 +16k2 ⇒ (AC)2 = 25k2 ⇒ AC =√25k2 ⇒ AC =±5k But side AC can’t be negative. So, AC = 5k
Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = BC = 4k and Hypotenuse = AC = 5k So, $\sin \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle θ = AB =3k Hypotenuse = AC =5k So, $\cos \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$ $\therefore \sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$ And $\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$
If $\cot \theta=\frac{1}{\sqrt{3}}$ verify that: $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5}$.Sol :We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \theta=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{3}}$Let, Side adjacent to angle θ =AB = 1k Side opposite to angle θ =AC = k√3 where, k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (AC)2 = (BC)2 ⇒ (1k)2 + (√3k)2 = (BC)2 ⇒ (BC)2 = 1 k2 +3 k2 ⇒ (BC)2 = 4 k2 ⇒ BC =√2 k2 ⇒ BC =±2k But side BC can’t be negative. So, BC = 2k Now, we will find the sin θ and cos θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = AC = k√3 and Hypotenuse = BC = 2k So$\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = AB =1k Hypotenuse = BC =2k So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$
If $\tan \theta=\frac{x}{y}$ find the value of x sin θ + y cos θ.Sol :
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ $\tan \theta=\frac{x}{y} \Rightarrow \frac{P}{B}=\frac{x}{y} \Rightarrow \frac{A B}{B C}=\frac{x}{y}$ Let, Side opposite to angle θ =AB = x Side adjacent to angle θ =BC = y where, k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (x)2 + (y)2 = (AC)2 ⇒ (AC)2 = x2+y2 ⇒ AC =√( x2+y2) Now, we will find the sin θ and cos θ We know that $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = BC = y and Hypotenuse = AC = √( x2+y2) So,$\cos \theta=\frac{B C}{A C}=\frac{y}{\sqrt{x^{2}+y^{2}}}$ And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ =AB = x And Hypotenuse =AC = √( x2+y2) So, $\sin \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}$ Now, x sin θ +y cos θ $=\mathrm{x}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}\right)+\mathrm{y}\left(\frac{\mathrm{y}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}\right)$ $=\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}}}$ = √( x2+y2)
Question 36
If $\sin \theta=\frac{3}{5}$, find the value of tan2θ + sinθ cosθ + cotθ.Given: $\sin \theta=\frac{3}{5}$We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypollyuse }}$ $\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$ Let, Perpendicular =AB =3k and Hypotenuse =AC =5k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle In right angled ∆ ABC, we have ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (BC)2 = (5k)2 ⇒ 9k2 + (BC)2 = 25k2 ⇒ (BC)2 = 25 k2 –9k2 ⇒ (BC)2 = 16k2 ⇒ BC =√16k2 ⇒ BC =±4k But side BC can’t be negative. So, BC = 4k Now, we have to find the value of cos θ and tan θ We know that, $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$ The side adjacent to angle θ or base = BC =4k Hypotenuse = AC =5k So,$\cos \theta=\frac{4 k}{5 k}=\frac{4}{5}$ Now, We know that, $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Perpendicular = AB =3k Base = BC =4k So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$ $\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$ Now, tan2 θ + sin θ cos θ + cot θ $=\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)+\left(\frac{4}{3}\right)$$=\left(\frac{9}{16}\right)+\left(\frac{13}{25}\right)+\left(\frac{4}{3}\right)$$=\frac{675+576+1600}{16 \times 25 \times 3}$$=\frac{2851}{1200}$
Question 37
If 4cot θ = 3, show that $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=7$.Sol : Given: $\cot \theta=\frac{3}{4}$We know that,$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{3}{4} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$ Let, Side adjacent to angle θ =AB = 3k The side opposite to angle θ =BC = 4k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (4k)2 = (AC)2 ⇒ (AC)2 = 9k2 +16k2 ⇒ (AC)2 = 25k2 ⇒ AC =√25k2 ⇒ AC =±5k But side AC can’t be negative. So, AC = 5k
Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle θ = BC = 4k and Hypotenuse = AC = 5k So, $\sin \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle θ = AB =3k Hypotenuse = AC =5k So, $\cos \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
If $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$, prove that $\mathrm{m} \sin \theta+\mathrm{n} \cos \theta=\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}$Sol : Given: $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$
Let, Perpendicular =AB =m and Hypotenuse =AC =√(m2 + n2) where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle In right angled ∆ ABC, we have ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (m)2 + (BC)2 = (√(m2 + n2))2 ⇒ m2 + (BC)2 = m2 + n2 ⇒ (BC)2 = m2 + n2 – m2 ⇒ (BC)2 = n2 ⇒ BC =√n2 ⇒ BC =±n But side BC can’t be negative. So, BC = n Now, we have to find the value of cos θ and tan θ
We know that, $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$ Side adjacent to angle θ or base = BC =n Hypotenuse = AC =√(m2 + n2) So, $\cos \theta=\frac{n}{\sqrt{m^{2}+n^{2}}}$
Now, LHS = m sin θ +n cosθ $=\mathrm{m}\left(\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}\right)+\mathrm{n}\left(\frac{\mathrm{n}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}\right)$ $=\frac{\mathrm{m}^{2}+\mathrm{n}^{2}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$ =√(m2 + n2) = RHS Hence Proved
Question 39
If $\cos \alpha=\frac{12}{13}$ show that $\sin \alpha(1-\tan \alpha)=\frac{35}{156}$Sol : We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$ $\cos \alpha=\frac{12}{13} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{12}{13} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{12}{13}$
Let, Base =BC = 12k Hypotenuse =AC = 13k Where, k ia any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (AB)2 + (12k)2 = (13k)2 ⇒ (AB)2 + 144k2 = 169k2 ⇒ (AB)2 = 169 k2 –144 k2 ⇒ (AB)2 = 25 k2 ⇒ AB =√25 k2 ⇒ AB =±5k But side AB can’t be negative. So, AB = 5k Now, we have to find sin α and tan α
We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle α = AB =5k And Hypotenuse = AC =13k So, $\sin \alpha=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Side opposite to angle α = AB =5k Side adjacent to angle α = BC =12k So, $\tan \alpha=\frac{5 \mathrm{k}}{12 \mathrm{k}}=\frac{5}{12}$ Now, LHS = sin α (1 – tan α)
$=\frac{5}{13}\left(1-\frac{5}{12}\right)$
$=\frac{5}{13}\left(\frac{12-5}{12}\right)$
$=\frac{35}{156}$ = RHS Hence Proved
Question 40
If $\mathrm{q} \cos \theta=\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}$, prove that q sin θ = p.Sol : Given : q cos θ = √(q2 – p2) $\Rightarrow \cos \theta=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}}$
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$ $\cos \theta=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}}$
Let, Base =BC = √(q2 – p2) Hypotenuse =AC = q Where, k ia any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (AB)2 + (√(q2 – p2))2 = (q)2 ⇒ (AB)2 + (q2 – p2) = q2 ⇒ (AB)2 = q2 – q2 + p2) ⇒ (AB)2 = p2 ⇒ AB =√p2 ⇒ AB =±p But side AB can’t be negative. So, AB = p Now, we have to find sin θ
We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ The side opposite to angle θ = AB =p And Hypotenuse = AC =q So, $\sin \theta=\left(\frac{p}{q}\right)$ Now, LHS = q sin θ $=\mathrm{q}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)$ = q = RHS Hence Proved
Question 41
If $\sin \theta=\frac{3}{5}$, show that : $\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}=-\frac{1}{5}$Sol : Given: $\sin \theta=\frac{3}{5}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$ Let, Perpendicular =AB =3k and Hypotenuse =AC =5k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle In right angled ∆ ABC, we have ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (3k)2 + (BC)2 = (5k)2 ⇒ 9k2 + (BC)2 = 25k2 ⇒ (BC)2 = 25 k2 –9k2 ⇒ (BC)2 = 16k2 ⇒ BC =√16k2 ⇒ BC =±4k But side BC can’t be negative. So, BC = 4k Now, we have to find the value of cos θ and tan θ
We know that, $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$ The side adjacent to angle θ or base = BC =4k Hypotenuse = AC =5k So, $\cos \theta=\frac{4 k}{5 k}=\frac{4}{5}$ Now, We know that, $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Perpendicular = AB =3k Base = BC =4k So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$ $\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$ Now, LHS $=\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$ $=\frac{\left(\frac{4}{5}\right)-\left(\frac{1}{\frac{3}{4}}\right)}{2\left(\frac{4}{3}\right)}$ $=\frac{\left(\frac{4}{5}\right)-\left(\frac{4}{3}\right)}{\left(\frac{8}{3}\right)}$ $=\frac{\frac{12-20}{15}}{\left(\frac{8}{3}\right)}$ $=\frac{\left(-\frac{8}{15}\right)}{\left(\frac{8}{3}\right)}$ $=-\frac{1}{5}$ = RHS Hence Proved
Question 42 A
Find the value ofcos A sin B + sin A. cos B, if sin A= 4/5 and cos B = 12/13. Sol :
Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$ To find: cos A sin B + sin A cos B As, we have the value of sin A and cos B but we don’t have the value of cos A and sin B So, First we find the value of cos A and sin B $\sin A=\frac{4}{5}$
We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin A=\frac{4}{5} \Rightarrow \frac{P}{H}=\frac{4}{5}$ Let, Side opposite to angle A = 4k and Hypotenuse = 5k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (P)2 + (B)2 = (H)2 ⇒ (4k)2 + (B)2 = (5)2 ⇒ 16 k2 + (B)2 = 25 k2 ⇒ (B)2 = 25 k2 –16 k2 ⇒ (B)2 = 9 k2 ⇒ B =√9 k2 ⇒ B =±3k [taking positive square root since, side cannot be negative] So, Base = 3k Now, we have to find the value of cos A We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle A =3k Hypotenuse =5k So, $\cos \mathrm{A}=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$ Now, we have to find the sin B
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ $\cos \mathbf{B}=\frac{12}{13} \Rightarrow \frac{B}{H}=\frac{12}{13}$ Let, Side adjacent to angle B =12k Hypotenuse =13k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (B)2 + (P)2 = (H)2 ⇒ (12k)2 + (P)2 = (13)2 ⇒ 144 k2 + (P)2 = 169 k2 ⇒ (P)2 = 169 k2 –144 k2 ⇒ (P)2 = 25 k2 ⇒ P =√25 k2 ⇒ P =±5k [taking positive square root since, side cannot be negative] So, Perpendicular = 5k Now, we have to find the value of sin B
We know that, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \mathrm{B}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$ Now, cos A sin B + sin A cos B
Putting the values of sin A, sin B cos A and Cos B, we get
Find the value ofsin A. cos B – cos A. sin B, if tan A= √3 and sin B = 1/2. Sol :
Given: tan A =√3 and $\sin B=\frac{1}{2}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \mathrm{B}=\frac{1}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{1}{2}$ Let, Side opposite to angle θ = 1k and Hypotenuse = 2k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AC)2 + (BC)2 = (AB)2 ⇒ (1k)2 + (BC)2 = (2k)2 ⇒ k2 + (BC)2 = 4k2 ⇒ (BC)2 = 4k2 –k2 ⇒ (BC)2 = 3 k2 ⇒ BC =√3k2 ⇒ BC =k√3 So, BC = k√3
Now, we have to find the value of cos B We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle B = BC =k√3 Hypotenuse = AB =2k So, $\cos \mathbf{B}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$
We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ Given: tan A = √3 $\Rightarrow \tan \mathrm{A}=\frac{\sqrt{3}}{1}$ $\tan \mathrm{A}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$ Let, The side opposite to angle A =BC = √3k The side adjacent to angle A =AB = 1k where k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (1k)2 + (√3k)2 = (AC)2 ⇒ (AC)2 = 1 k2 +3 k2 ⇒ (AC)2 = 4 k2 ⇒ AC =√2 k2 ⇒ AC =±2k But side AC can’t be negative. So, AC = 2k Now, we will find the sin A and cos A $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle A = BC = k√3 and Hypotenuse = AC = 2k So, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ The side adjacent to angle A = AB =1k Hypotenuse = AC =2k
So, $\cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$ Now, sin A. cos B – cos A. sin B
Putting the values of sin A, sin B cos A and Cos B, we get
Find the value ofsin A. cos B + cos A. sin B. if $\tan \mathrm{A}=\frac{1}{\sqrt{3}}$and tan B = √3. Sol : Given:
$\tan \mathrm{A}=\frac{1}{\sqrt{3}}$ $\tan \mathrm{A}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{1}{\sqrt{3}}$ Let, Side opposite to angle A =BC = 1k Side adjacent to angle A =AB = k√3 where, k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (√3k)2 + (1k)2 = (AC)2 ⇒ (AC)2 = 1 k2 +3 k2 ⇒ (AC)2 = 4 k2 ⇒ AC =√2 k2 ⇒ AC =±2k But side AC can’t be negative. So, AC = 2k
Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle A = BC = k and Hypotenuse = AC = 2k So, $\operatorname{Sin} \mathbf{A}=\frac{B C}{A C}=\frac{1 k}{2 k}=\frac{1}{2}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle A = AB =k√3 Hypotenuse = AC =2k So, $\cos \mathbf{A}=\frac{A B}{B C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$ Now,
Given: tan B = √3 $\Rightarrow \tan \mathrm{B}=\frac{\sqrt{3}}{1}$ $\tan \mathrm{B}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$ Let, Side opposite to angle B =AC = √3k Side adjacent to angle B =AB = 1k where, k is any positive integer Firstly we have to find the value of BC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (AC)2 = (BC)2 ⇒ (1k)2 + (√3k)2 = (BC)2 ⇒ (BC)2 = 1 k2 +3 k2 ⇒ (BC)2 = 4 k2 ⇒ BC =√2 k2 ⇒ BC =±2k But side BC can’t be negative. So, BC = 2k
Now, we will find the sin B and cos B $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle B = AC = k√3 and Hypotenuse = BC = 2k So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle B = AB =1k Hypotenuse = BC =2k So, $\cos \mathbf{B}=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$ Now, sin A. cos B + cos A. sin B
Putting the values of sin A, sin B cos A and Cos B, we get
Find the value of$\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$, if sin A = $\frac{1}{\sqrt{2}}$and cos $\mathrm{B}=\frac{\sqrt{3}}{2}$ Sol : Given $: \sin \mathrm{A}=\frac{1}{\sqrt{2}}$ and $\cos \mathrm{B}=\frac{\sqrt{3}}{2}$ $\sin \mathrm{A}=\frac{1}{\sqrt{2}}$
We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \mathrm{A}=\frac{1}{\sqrt{2}} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{1}{\sqrt{2}}$
Let, Side opposite to angle A = k and Hypotenuse = k√2 where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (P)2 + (B)2 = (H)2 ⇒ (k)2 + (B)2 = (k√2)2 ⇒ k2 + (B)2 = 2k2 ⇒ (B)2 = 2k2 – k2 ⇒ (B)2 = k2 ⇒ B =√k2 ⇒ B =±k [taking positive square root since, side cannot be negative] So, Base = k Now, we have to find the value of tan A
We know that, $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ So, $\tan \mathrm{A}=\frac{\mathrm{k}}{\mathrm{k}}=1$
Now, we have to find the tan B We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ $\cos \mathbf{B}=\frac{\sqrt{3}}{2} \Rightarrow \frac{B}{H}=\frac{\sqrt{3}}{2}$ Let, Side adjacent to angle B =k√3 Hypotenuse =2k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (B)2 + (P)2 = (H)2 ⇒ (k√3)2 + (P)2 = (2k)2 ⇒ 3k2 + (P)2 = 4k2 ⇒ (P)2 = 4k2 –3 k2 ⇒ (P)2 = k2 ⇒ P =√k2 ⇒ P =±k [taking positive square root since, side cannot be negative] So, Perpendicular = k Now, we have to find the value of sin B We know that, $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$ So, $\tan \mathrm{B}=\frac{\mathrm{k}}{\mathrm{k} \sqrt{3}}=\frac{1}{\sqrt{3}}$
Now, $\frac{\tan A+\tan B}{1-\tan A \tan B}$ $\Rightarrow \frac{(1)+\left(\frac{1}{\sqrt{3}}\right)}{1-(1)\left(\frac{1}{\sqrt{3}}\right)}$
Now, multiply and divide by the conjugate of √3 – 1, we get $\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$ $\Rightarrow \frac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$ [∵ (a – b)(a+b) = (a2 – b2)]
$\Rightarrow \frac{3+1+2 \sqrt{3}}{3-1}$
$\Rightarrow \frac{4+2 \sqrt{3}}{2}$
⇒ 2+√3
Question 42 E
Find the value ofsec A. tan A+tan2A – cosec A, if tan A =2 Sol : Given: tan A = 2 ⇒ tan2A = 4 We know that, sec2 A = 1+ tan2A ⇒ sec2 A = 1 + 4 ⇒ sec2 A = 5 ⇒ sec A =√5 $\Rightarrow \cos A=\frac{1}{\sqrt{5}}$
Now, we know that tan A$=\frac{\sin A}{\cos A}$ $\Rightarrow 2=\frac{\sin A}{\frac{1}{\sqrt{5}}}$ ⇒ 2 =√5 sin A
Find the value of$\frac{1}{\tan \mathrm{A}}+\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}$, if cosec A = 2 Sol : Given: cosec A =2 Now, we have to find $\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$ First, we simplify the above given trigonometry equation, we get $\frac{1}{\frac{\sin A}{\cos A}}+\frac{\sin A}{1+\cos A}$ $\Rightarrow \frac{\cos A}{\sin A}+\frac{\sin A}{1+\cos A}$
If $\sin \mathrm{B}=\frac{1}{2}$, prove that : 3 cos B – 4cos3 B = 0Sol : Given: $\sin B=\frac{1}{2}$
We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin B=\frac{1}{2} \Rightarrow \frac{P}{H}=\frac{1}{2} \Rightarrow \frac{A B}{A C}=\frac{1}{2}$ Let, Perpendicular =AB =k and Hypotenuse =AC =2k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle In right angled ∆ ABC, we have ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (k)2 + (BC)2 = (2k)2 ⇒ k2 + (BC)2 = 4k2 ⇒ (BC)2 = 4k2 –k2 ⇒ (BC)2 = 3k2 ⇒ BC =√3k2 ⇒ BC =k√3 So, BC = k√3 Now, we have to find the value of cos B
We know that, $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$ Side adjacent to angle B or base = BC = k√3 Hypotenuse = AC =2k So, $\cos \mathrm{B}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$ Now, LHS = 3 cos B – 4cos3 B
If $\sec \theta=\frac{5}{4}$, prove that : $\frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\sin \theta}{\sec \theta}$Sol : Given: $\sec \theta=\frac{5}{4}$
We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$ $\operatorname{Sec} \theta=\frac{5}{4} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{5}{4} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{4}$ Let, BC = 4k and AC = 5k where, k is any positive integer. In right angled ∆ABC, we have (AB)2 + (BC)2 = (AC)2 [by using Pythagoras theorem] ⇒ (AB)2 + (4k)2 = (5k )2 ⇒ (AB)2 + 16k2 = 25k2 ⇒ (AB)2 = 25k2 – 16k2 ⇒ (AB)2 = 9k2 ⇒ AB = √9k2 ⇒ AB =±3k [taking positive square root since, side cannot be negative] Now, we have to find the value of other trigonometric ratios. We, know that
$\cot B=\frac{12}{5}$, prove that : tan2B – sin2 B=sin4 B sec2 B.Sol :
We know that, $\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$ Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$ $\cot B=\frac{12}{5} \Rightarrow \frac{B}{P}=\frac{12}{5} \Rightarrow \frac{A B}{B C}=\frac{12}{5}$ Let, Side adjacent to angle B =AB = 12k Side opposite to angle B =BC = 5k where, k is any positive integer Firstly we have to find the value of AC. So, we can find the value of AC with the help of Pythagoras theorem ⇒ (AB)2 + (BC)2 = (AC)2 ⇒ (12k)2 + (5k)2 = (AC)2 ⇒ (AC)2 = 144 k2 +25 k2 ⇒ (AC)2 = 169 k2 ⇒ AC =√169 k2 ⇒ AC =±13k But side AC can’t be negative. So, AC = 13k
Now, we will find the sin θ $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Side opposite to angle B = BC = 5k and Hypotenuse = AC = 13k So, $\sin \mathrm{B}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$ Now, we know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle B = AB =12k Hypotenuse = AC =13k So, $\cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 \mathrm{k}}{13 \mathrm{k}}=\frac{12}{13}$ $\tan \mathrm{B}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$ $\sec \mathrm{B}=\frac{1}{\cos \mathrm{B}}=\frac{1}{\frac{12}{13}}=\frac{13}{12}$
If $\cos \theta=\frac{\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}$, prove that $\left(\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}= \frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}-\mathrm{q}}}$Sol : Given: $\cos \theta=\frac{q}{\sqrt{p^{2}+q^{2}}}$ Now, squaring both the sides, we get
Now, solving LHS $=\left(\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$ Putting the value of p2 in the above equation, we get
∴ LHS = RHS Hence Proved Question 45 In the given figure, BC = 15 cm and sin B = 4/5, show that $\tan ^{2} \mathrm{B}-\frac{1}{\cos ^{2} \mathrm{B}}=-1$
Sol : Given: BC =15cm and $\sin B=\frac{4}{5}$ We know that, $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$ Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ $\sin \mathrm{B}=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{4}{5}$ Let, Side opposite to angle B = 4k and Hypotenuse = 5k where, k is any positive integer So, by Pythagoras theorem, we can find the third side of a triangle ⇒ (AC)2 + (BC)2 = (AB)2 ⇒ (4k)2 + (BC)2 = (5)2 ⇒ 16k2 + (BC)2 = 25k2 ⇒ (BC)2 = 25 k2 –16 k2 ⇒ (BC)2 = 9 k2 ⇒ BC =√9 k2 ⇒ BC =±3k But side BC can’t be negative. So, BC = 3k Now, we have to find the value of cos B and tan B
We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$ Side adjacent to angle B = BC =3k Hypotenuse = AB =5k So, $\cos B=\frac{3 k}{5 k}=\frac{3}{5}$ Now, tan B We know that, $\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$ side opposite to angle B = AC =4k Side adjacent to angle B = BC =3k So, $\tan \mathrm{B}=\frac{4 \mathrm{k}}{3 \mathrm{k}}=\frac{4}{3}$
In the given figure, find 3 tan θ – 2 sin α + 4 cos α.
Sol : First of all, we find the value of RS In right angled ∆RQS, we have (RQ)2 + (QS)2 = (RS)2 ⇒ (8)2 + (6)2 = (RS)2 ⇒ 64 + 36 = (RS)2 ⇒ RS =√100 ⇒ RS =±10 [taking positive square root, since side cannot be negative] ⇒ RS =10
In the given figure ΔABC is right angled at B and BD is perpendicular to AC. Find (i) cos θ, (ii) cot α.
Sol : Firstly, we find the value of AC In right angled ∆ABC (AB)2 + (BC)2 = (AC)2 ⇒ (12)2 + (5)2 = (AC)2 ⇒ 144+25 =(AC)2 ⇒ (AC)2 =169 ⇒ AC =√169 ⇒ AC =±13 ⇒ AC =13 [taking positive square root since, side cannot be negative] (i) $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{12}{13}$ (ii) $\cot \alpha=\frac{\text { Base }}{\text { Perpendicular }}=\frac{12}{5}$
If 4 cos θ + 3 sin θ = 5, find the value of tan θ.Sol : Given : 4 cos θ+ 3 sin θ = 5 Squaring both the sides, we get ⇒ (4 cos θ+ 3 sin θ)2 = 25 ⇒ 16 cos2 θ + 9 sin2 θ + 2(4cos θ)(3sin θ)= 25 [∵ (a + b)2 =a2 +b2 +2ab] ⇒ 16 cos2 θ + 9 sin2 θ + 24 cosθ sinθ = 25
If 7 sin A + 24 cos A = 25, find the value of tan A.Sol : Given : 7 sin A + 24 cos A = 25 Squaring both the sides, we get ⇒ (7 sin A + 24 cos A)2 = 625 ⇒ 49 sin2 A +576 cos2 A + 2(7sin A) (24cos A) = 625 [∵ (a + b)2 =a2 +b2 +2ab] ⇒ 49 sin2 A +576 cos2 A + 336 cosA sinA = 625
Divide by cos2 θ, we get $\Rightarrow \frac{49 \sin ^{2} \mathrm{A}}{\cos ^{2} \mathrm{A}}+\frac{576 \cos ^{2} \mathrm{A}}{\cos ^{2} \mathrm{A}}+\frac{336 \cos \mathrm{A} \sin \mathrm{A}}{\cos ^{2} \mathrm{A}}=\frac{625}{\cos ^{2} \mathrm{A}}$
⇒ 49tan2 A +576+ 336 tanA = 625sec2 A ⇒ 49tan2 A +576+ 336 tanA = 625(1 + tan2 A) [∵ 1+ tan2θ = sec2 θ] ⇒ 49tan2 A +576+ 336 tanθA = 625+625 tan2 A ⇒ 576tan2 A – 336tanA + 49 = 0 ⇒ 576tan2 A – 168 tanA – 168 tanA +49 = 0 ⇒ 24tanθ (24tan A – 7) – 7(24tan A – 7) = 0 ⇒ (24tan A – 7)2 = 0 $\Rightarrow \tan \mathrm{A}=\frac{7}{24}$
Question 52
If 9 sin θ + 40 cos θ= 41, find the value of cos θ and cosec θSol : Given: 9 sin θ + 40 cos θ= 41 ⇒ 9sinθ = 41 – 40 cosθ …(i) Squaring both sides, we get ⇒ 81sin2 θ = 1681+1600 cos2 θ – 2(41) (40cos θ) [∵ (a – b)2 =a2 +b2 –2ab] ⇒ 81 (1– cos2 θ) =1681+1600 cos2 θ – 3280cosθ ⇒ 81 – 81cos2 θ = 1681 +1600cos2 θ – 3280 cosθ ⇒ 1681cos2 θ –3280cos θ +1600 = 0 ⇒ (41)2 cos2 θ – 2(41) (40cos θ) + (40)2 = 0 ⇒ (41cos θ – 40 )2 = 0 $\Rightarrow \cos \theta=\frac{40}{41}$Now, putting the value of cos θ in Eq. (i), we get
If tan A + sec A = 3, find the value of sin A.Sol : tan A + sec A = 3 ⇒ tanA = 3 – secA Squaring both the sides, we get ⇒ tan2 A =(3 – secA)2 ⇒ tan2 A = 9 + sec2A – 6sec A ⇒ sec2 A – 1 = 9 + sec2A – 6sec A [∵ 1+ tan2 A = sec2 A] ⇒ –1 – 9 = –6secA ⇒ – 10 = –6sec A $\Rightarrow \sec A=\frac{10}{6}$ $\Rightarrow \frac{1}{\cos A}=\frac{5}{3}\left[\because \sec A=\frac{1}{\cos A}\right]$ $\Rightarrow \cos A=\frac{3}{5}$
Now, tan A + sec A = 3 $\Rightarrow \frac{\sin A}{\cos A}+\frac{1}{\cos A}=3\left[\because \tan A=\frac{\sin A}{\cos A}\right]$ $\Rightarrow \frac{\sin A}{\cos A}=\frac{3 \cos A-1}{\cos A}$ ⇒ sin A = 3cosA – 1
$\Rightarrow \sin A=3\left(\frac{3}{5}\right)-1$
$\Rightarrow \sin A=\left(\frac{9-5}{5}\right)$
$\Rightarrow \sin A=\left(\frac{4}{5}\right)$
Question 54
If cosec A + cot A = 5, find the value of cos A.Sol : cosec A + cot A = 5 ⇒ cotA = 5 – cosecA Squaring both the sides, we get ⇒ cot2 A =(5 – cosecA)2 ⇒ cot2 A = 25 + cosec2A – 10cosec A ⇒ cosec2 A – 1 = 25 + cosec2A – 10cosec A [∵ 1+ cot2 A = cosec2 A] ⇒ –1 – 25 = –10cosecA ⇒ – 26 = –10cosec A $\Rightarrow \operatorname{cosec} A=\frac{26}{10}$ $\Rightarrow \frac{1}{\sin A}=\frac{13}{5}\left[\because \operatorname{cosec} A=\frac{1}{\sin A}\right]$ $\Rightarrow \sin \mathrm{A}=\frac{5}{13}$
Now, cosec A + cot A = 5 $\Rightarrow \frac{1}{\sin A}+\frac{\cos A}{\sin A}=5\left[\because \cot A=\frac{\cos A}{\sin A}\right]$ $\Rightarrow \frac{13}{5}+\frac{\cos A}{\frac{5}{13}}=5$
$\Rightarrow \frac{13}{5}+\frac{13 \cos A}{5}=5$
$\Rightarrow \frac{13 \cos A}{5}=5-\frac{13}{5}$
$\Rightarrow \frac{13 \cos A}{5}=\frac{25-13}{5}$
$\Rightarrow \cos A=\frac{12}{13}$
Question 55
If tan θ + sec θ = x, show that $\sin \theta=\frac{x^{2}-1}{x^{2}+1}$Sol : tan θ+ sec θ = x ⇒ tan θ = x – sec θ Squaring both sides, we get ⇒ tan2 θ =(x – secθ)2 ⇒ tan2 θ = x2 + sec2θ – 2xsec θ ⇒ sec2 θ – 1 = x2 + sec2θ – 2xsec θ [∵ 1+ tan2 A = sec2 A] ⇒ –1 – x2 = –2xsecθ $\Rightarrow \sec \theta=\frac{1+x^{2}}{2 x}$ Now, tan θ = x – sec θ $\Rightarrow \frac{\sin \theta}{\cos \theta}=x-\sec \theta$ $\Rightarrow \sin \theta\left(\frac{1+x^{2}}{2 x}\right)=x-\left(\frac{1+x^{2}}{2 x}\right)$ $\Rightarrow \sin \theta\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)=\left(\frac{2 \mathrm{x}^{2}-1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)$ $\Rightarrow \sin \theta\left(\frac{1+x^{2}}{2 x}\right)=\left(\frac{x^{2}-1}{2 x}\right)$ $\Rightarrow \sin \theta=\frac{x^{2}-1}{x^{2}+1}$ = RHS Hence Proved
Question 56
If cos θ +sin θ=1, prove that cos θ – sin θ = ± 1Sol : Using the formula, (a+b)2 + (a – b)2 = 2(a2+b2) ⇒ (cos θ +sin θ)2 + (cos θ – sin θ)2 = 2(cos2θ + sin2 θ) ⇒ 1 + (cos θ – sin θ)2 = 2(1) ⇒ (cos θ – sin θ)2 = 2 –1 ⇒ (cos θ – sin θ)2 = 1 ⇒ (cos θ – sin θ) =√1 ⇒ (cos θ – sin θ) = ±1
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