KC Sinha: Exercise 4.2 - Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities

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Question 1 

Find the value of the following :

(i) sin 30^o + cos 60^o
(ii) sin² 45^o+cos²45^o
(iii) sin 30^o + cos 60^o – tan45^o
(iv) $\sqrt{1+\tan ^{2} 60^{\circ}}$
(v) tan 60^o x cos30^o
Sol :
(i) sin 30^o + cos 60^o
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}>\cos \left(60^{\circ}\right)=\frac{1}{2}$

So,
sin(30^o) + cos(60^o)
$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)$
=1

(ii) sin² 45^o+cos²45^o
We know that,
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

So, sin² 45^o+cos²45^o
$=\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)$
=1

(iii) sin 30^o + cos 60^o – tan45^o
$\sin \left(30^{\circ}\right)=\frac{1}{2}$
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
tan(45^o)=1

So,
sin 30^o + cos 60^o – tan 45^o
$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)-1$
$=\frac{1+1-2}{2}$
=0

(iv) $\sqrt{1+\tan ^{2} 60^{\circ}}$
We know that
tan(60^o) = √3
So,
$=\sqrt{1+\tan ^{2} 60^{\circ}}$
$=\sqrt{1+(\sqrt{3})^{2}}$
$=\sqrt{1+3}$
=√4
= 2
(v) tan 60^o × cos30^o
tan(60^o) = √3
$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
So,
tan 60^o × cos30^o
$=\sqrt{3} \times \frac{\sqrt{3}}{2}$
$=\frac{3}{2}$

---

Question 2 

If θ = 45°, find the value of

(i) $\tan ^{2} \theta+\frac{1}{\sin ^{2} \theta}$
(ii) cos² θ - sin² θ
Sol :
(i) $\tan ^{2} \theta+\frac{1}{\sin ^{2} \theta}$
Given θ =45°
We know that,
tan(45^o) = 1
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$=(1)^{2}+\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}}$
= 1+ 2
= 3
(ii) cos² θ – sin² θ
Given θ = 45°
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
So, cos² 45° – sin² 45°
$=\left(\frac{1}{\sqrt{2}}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}$
= 0

---

Question 3 A 

Find the numerical value of the following :

sin45°.cos45° – sin²30°.
Sol :
We know that,
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Now, putting the values
$=\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{2}\right)^{2}$
$=\left(\frac{1}{2}\right)-\left(\frac{1}{4}\right)$
$=\left(\frac{1}{4}\right)$

---

Question 3 B 

Find the numerical value of the following :

$\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}$
Sol :
We know that,
$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
tan (60^°) = √3

Now putting the values;

$=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\frac{1}{2}}$

$=\frac{\frac{\sqrt{3}}{1+\sqrt{3}}}{2}$

$=\sqrt{3} \times \frac{2}{1+\sqrt{3}}$

$=\frac{2 \sqrt{3}}{1+\sqrt{3}}$

Multiplying and dividing by the conjugate of (1+√3)
$=\frac{2 \sqrt{3}}{1+\sqrt{3}} \times \frac{1-\sqrt{3}}{1-\sqrt{3}}$
$=\frac{2 \sqrt{3}-6}{(1)^{2}-(\sqrt{3})^{2}}$ [∵(a)² – (b)² = (a+b)(a-b)]
$=\frac{2 \sqrt{3}-6}{-2}$

Multiplying and dividing by (-2)
= 3 - √3

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Question 3 C 

Find the numerical value of the following :

$\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 30^{\circ}}$
Sol :
We know that,
$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$
$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
tan (60^o) = √3

Now putting the values;

$=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}$

$=\frac{\sqrt{3}}{\sqrt{3}}$

= 1

---

Question 3 D 

Find the numerical value of the following :

$\frac{4}{\sin ^{2} 60^{\circ}}+\frac{3}{\cos ^{2} 60^{\circ}}$
Sol :
We know that

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Now putting the value, we get
$=\frac{4}{\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{\left(\frac{1}{2}\right)^{2}}$
$=4 \times\left(\frac{2}{\sqrt{3}}\right)^{2}+3 \times(2)^{2}$
$=4\left(\frac{4}{3}\right)+3 \times 4$
$=\frac{16}{3}+12$
$=\frac{16+36}{3}$
$=\frac{52}{3}$

---

Question 3 E 

Find the numerical value of the following :

sin² 60° – cos² 60°
Sol :
We know that,

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Now putting the value;
$=\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$
$=\left(\frac{3}{4}\right)-\left(\frac{1}{4}\right)$
$=\frac{1}{2}$

---

Question 3 F 

Find the numerical value of the following :

4sin² 30° + 3 tan 30° – 8 sin 45° cos 45°
Sol :
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Now putting the value, we get
$=4 \times\left(\frac{1}{2}\right)^{2}+3 \times\left(\frac{1}{\sqrt{3}}\right)^{2}-8 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$=4 \times \frac{1}{4}+3 \times \frac{1}{3}-8 \times \frac{1}{2}$
= 1 + 1 – 4
= -2

---

Question 3 G 

Find the numerical value of the following :

2sin²30° – 3cos² 45° + tan² 60ׄ°
Sol :
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Tan (60^o) = √3
Now putting the value;
$=2 \times\left(\frac{1}{2}\right)^{2}-3 \times\left(\frac{1}{\sqrt{2}}\right)^{2}+(\sqrt{3})^{2}$
$=2 \times \frac{1}{4}-3 \times \frac{1}{2}+3$
$=\frac{1}{2}-\frac{3}{2}+3$

$=\frac{1-3+6}{2}$

$=\frac{4}{2}$

=2

---

Question 3 H 

Find the numerical value of the following :

sin 90° + cos 0° + sin 30° + cos 60°
Sol :
We know that,
Sin (90^o) = 1
Cos (0^o) = 1

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Now putting the value;

$=1+1+\frac{1}{2}+\frac{1}{2}$

$=\frac{2+2+1+1}{2}$

$=\frac{6}{2}$

= 3

---

Question 3 I 

Find the numerical value of the following :

sin 90° – cos 0° + tan 0° + tan 45°
Sol :
We know that
Sin (90^o) = 1
Cos (0^o) = 1
Tan(0^o) = 0
Tan(45^o) = 1
Now putting the value, we get
= 1 – 1 + 0 + 1
= 1

---

Question 3 J 

Find the numerical value of the following :

$\cos ^{2} 0^{\circ}+\tan ^{2} \frac{\pi}{4}+\sin ^{2} \frac{\pi}{4}$, where π = 180°
Sol :
We know that
Cos (0^o) = 1
Tan (45^o) = 1 $\left[\frac{\pi}{4}=\frac{180^{\circ}}{4}=45^{\circ}\right]$
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}\left[\frac{\pi}{4}=\frac{180^{\circ}}{4}=45^{\circ}\right]$

Now putting the values;
$=(1)^{2}+(1)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=1+1+\frac{1}{2}$
$=\frac{2+2+1}{2}$
$=\frac{5}{2}$

---

Question 3 K 

Find the numerical value of the following :

$\frac{\cos 60^{\circ}}{\sin ^{2} 45^{\circ}}-3 \cot 45^{\circ}+2 \sin 90^{\circ}$
Sol :
We know that,
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
Cot (45^o) = 1
Sin (90^o) = 1
Now putting the values, we get
$=\frac{\frac{1}{2}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}-3(1)+2(1)$
= 1-3+2
=0

---

Question 3 L 

Find the numerical value of the following :

$\frac{4}{\tan ^{2} 60^{\circ}}+\frac{1}{\cos ^{2} 30^{\circ}}-\sin ^{2} 45^{\circ}$
Sol :
We can write the above equation as:
= 4 cot² 60^o + sec² 30^o – sin² 45^o …(a) $\left[\because \cos \theta=\frac{1}{\sec \theta}\right.$ and $\left.\tan \theta=\frac{1}{\cot \theta}\right]$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cot \left(60^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\operatorname{Sec}\left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

Now putting the values in (a);
$=4\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}$

$=4 \times \frac{1}{3}+\frac{4}{3}-\frac{1}{2}$

$=\frac{8+8-3}{6}$

$=\frac{13}{6}$

---

Question 3 M 

Find the numerical value of the following :

cos60° . cos 30° – sin 60° . sin 30°
Sol :
We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Now putting the values, we get

$=\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}$

$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

= 0

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Question 3 N 

Find the numerical value of the following :

$\frac{4\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)}{\tan ^{2} 30^{\circ}+\cos ^{2} 90^{\circ}}$
Sol :
We know that,

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

cos(90^o) = 0

Now putting the values;
$=4 \times \frac{\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}}{\left(\frac{1}{\sqrt{3}}\right)^{2}-(0)^{2}}$
$=4 \times \frac{\frac{3}{4}-\frac{1}{2}}{\frac{1}{3}}$
$=4 \times \frac{1}{4} \times 3$
= 3

---

Question 4 A 

Evaluate the following :

sin30°.cos45° + cos30°.sin45°
Sol :

We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Now putting the values, we get

$=\frac{1}{2} \times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}$

$=\frac{1}{2 \sqrt{2}}+\frac{\sqrt{3}}{2 \sqrt{2}}$

$=\frac{1+\sqrt{3}}{2 \sqrt{2}}$

---

Question 4 B 

Evaluate the following :

cosec²30°.tan²45° – sec²60°
Sol :
We know that
cosec (30^o) = 2
Tan(45^o) = 1
sec (60 ^o) = 2
Now putting the values;
= (2)² × (1)² - (2)²
= 4 – 4
= 0

---

Question 4 C 

Evaluate the following :

2sin²30°.tan60° – 3cos²60°.sec²30°
Sol :
We know that
$\sin \left(30^{\circ}\right)=\frac{1}{2}$
tan (60^o) = √3

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\sec \left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

Now putting the values;

$=2 \times\left(\frac{1}{2}\right)^{2} \times(\sqrt{3})-\left(3 \times\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{\sqrt{3}}\right)^{2}\right)$

$=2 \times \frac{1}{4} \times(\sqrt{3})-\left(3 \times \frac{1}{4} \times \frac{4}{3}\right)$

$=\frac{\sqrt{3}}{2}-1$

$=\frac{\sqrt{3}-2}{2}$

---

Question 4 D 

Evaluate the following :

tan60° . cosec²45° + sec²60°.tan45°
Sol :
We know that
tan (60^o) = √3
cosec (45^o) = √2
sec (60 ^o) = 2
tan(45^o) = 1
Now putting the values;
= (√3) × (√2)² + (2)² × (1)
= 2√3 +4
=2 (√3 + 2)

---

Question 4 E 

Evaluate the following :

tan30°.sec45° + tan60°.sin30°
Sol :
We know that
$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$
sec (45^o) = √2
tan (60^o) = √3
$\sec \left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

Now putting the values, we get
$=\frac{1}{\sqrt{3}} \times \sqrt{2}+\sqrt{3} \times \frac{2}{\sqrt{3}}$
$=\frac{\sqrt{2}}{\sqrt{3}}+2$
$=2+\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$=2+\frac{\sqrt{6}}{3}$

---

Question 4 F 

Evaluate the following :

cos30°.cos45° – sin30°.sin45°
Sol :
We know that

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Now putting the values, we get
$=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times \frac{1}{\sqrt{2}}$
$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

Multiplying and dividing by (√2), we get

$=\frac{\sqrt{3}-1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{\sqrt{2}(\sqrt{3}-1)}{4}$

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Question 4 G 

Evaluate the following :

$\frac{4}{3} \tan ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}-3 \cos ^{2} 60^{\circ}+\frac{3}{4} \tan ^{2} 60^{\circ}-2 \tan ^{2} 45^{\circ}$
Sol :
We know that
$\begin{aligned} \tan \left(30^{\circ}\right) &=\frac{1}{\sqrt{3}} \\ \sin \left(60^{\circ}\right) &=\frac{\sqrt{3}}{2} \\ \cos \left(60^{\circ}\right) &=\frac{1}{2} \end{aligned}$
tan (60^o) = √3
tan(45^o) = 1

Now putting the values;
$=\left(\frac{4}{3} \times\left(\frac{1}{\sqrt{3}}\right)^{2}\right)+\left[\left(\frac{\sqrt{3}}{2}\right)^{2}\right]-\left[3 \times\left(\frac{1}{2}\right)^{2}\right]+\left[\frac{3}{4}(\sqrt{3})^{2}\right]-\left[2 \times(1)^{2}\right]$
$=\left[\frac{4}{3} \times \frac{1}{3}\right]+\left[\frac{3}{4}\right]-\left[3 \times \frac{1}{4}\right]+\left[\frac{3}{4} \times 3\right]-[2 \times(1)]$

$=\frac{4}{9}+\frac{3}{4}-\frac{3}{4}+\frac{9}{4}-2$

$=\frac{16+27-27+81-72}{36}$

$=\frac{25}{36}$

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Question 4 H 

Evaluate the following :

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$
Sol :
We know that
tan (60^o) = √3

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\sec \left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

cos(90^o) = 0
cosec (30^o) = 2
sec (60 ^o) = 2
cot (30^o) = √3

Now putting the values, we get
$=\frac{(\sqrt{3})^{2}+\left[4 \times\left(\frac{1}{\sqrt{2}}\right)^{2}\right]+\left[3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}\right]+\left[5 \times(0)^{2}\right]}{(2)+(2)-(\sqrt{3})^{2}}$
$=\frac{(3)+\left[4 \times \frac{1}{2}\right]+\left[3 \times \frac{4}{3}\right]+[5 \times 0]}{2+2-3}$
$=\frac{(3)+[2]+[4]+[0]}{2+2-3}$
= 9

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Question 4 I 

Evaluate the following :

$\frac{5 \sin ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}-4 \tan ^{2} 30^{\circ}}{2 \sin 30^{\circ} \cdot \cos 30^{\circ}+\tan 45^{\circ}}$
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

tan (45^o) = 1

Now putting the values, we get
$=\frac{\left[5 \times\left(\frac{1}{2}\right)^{2}\right]+\left(\frac{1}{\sqrt{2}}\right)^{2}-\left[4 \times\left(\frac{1}{\sqrt{3}}\right)^{2}\right]}{2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+(1)}$
$=\frac{\left(\frac{5}{4}\right)+\left(\frac{1}{2}\right)-\left(\frac{4}{3}\right)}{\left(\frac{\sqrt{3}}{2}\right)+1}$
$=\frac{\left(\frac{15+6-16}{12}\right)}{\left(\frac{\sqrt{3}+2}{2}\right)}$
$=\frac{5}{12} \times \frac{2}{\sqrt{3}+1}$
$=\frac{5}{6} \times \frac{1}{\sqrt{3}+2}$

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Question 5 A 

Prove the following :

$\frac{(1-\cos B)(1+\cos B)}{(1-\sin B)(1+\sin B)}=\frac{1}{3}$ When B = 30°
Sol :
Solving, L.H.S.
$=\frac{(1)^{2}-(\cos B)^{2}}{(1)^{2}-(\sin B)^{2}}$ [(a)² – (b)² = (a+b)(a-b)]
$=\frac{1-\cos ^{2} B}{1-\sin ^{2} B}$

We know that,
$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Putting the values, we get
$=\frac{1-\left(\frac{\sqrt{3}}{2}\right)^{2}}{1-\left(\frac{1}{2}\right)^{2}}$
$=\frac{1-\frac{3}{4}}{1-\frac{1}{4}}$
$=\frac{4-3}{4-1}$
$=\frac{1}{3}$
=R.H.S.
Hence Proved

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Question 5 B 

Prove the following :

$\frac{(1-\cos \alpha)(1+\cos \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}=3$ When α =60°
Sol :
Solving, L.H.S.
$=\frac{(1)^{2}-(\cos \alpha)^{2}}{(1)^{2}-(\sin \alpha)^{2}}$ [(a)² – (b)² = (a+b)(a-b)]
$=\frac{1-\cos ^{2} \alpha}{1-\sin ^{2} \alpha}$

We know that

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

Putting the values, we get
$=\frac{1-\left(\frac{1}{2}\right)^{2}}{1-\left(\frac{\sqrt{3}}{2}\right)^{2}}$

$=\frac{1-\frac{1}{4}}{1-\frac{3}{4}}$

$=\frac{4-1}{4-3}$
= 3 = R.H.S.

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Question 5 C 

Prove the following :

cos(A – B) = cos A. cos B + sinA . sin B if A=B=60^o
Sol :
Solving, L.H.S.
= cos (60^o – 60^o) [Putting the value A=B=60^o]
= cos (0^o)
= 1
Solving, R.H.S.
= cos (60^o) × cos (60^o) + sin (60^o) × sin (60^o) [Putting the value A=B=60^o]
= cos²(60^o) + sin²(60^o)

We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$=\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}$

$=\frac{1}{4}+\frac{3}{4}$

$=\frac{1+3}{4}$

= 1
∴ LHS = RHS
Hence Proved

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Question 5 D 

Prove the following :

4(sin⁴30° + cos⁴ 60°) – 3(cos² 45° – sin²90°) = 2
Sol :
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Sin (90^o) = 1

= 4[{(sin 30^o)²}² + {(cos 60^o)²}²] – 3[(cos 45^o)² - (sin 90^o)²]

Putting the values
$=4 \times\left[\left\{\left(\frac{1}{2}\right)^{2}\right\}^{2}+\left\{\left(\frac{1}{2}\right)^{2}\right\}^{2}\right]-3\left[\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right]$
$=4 \times\left[\left\{\frac{1}{4}\right\}^{2}+\left\{\frac{1}{4}\right\}^{2}\right]-3\left[\frac{1}{2}-1\right]$
$=4 \times\left[\frac{1}{16}+\frac{1}{16}\right]-3\left[-\frac{1}{2}\right]$
$=4 \times\left[\frac{1}{8}\right]-3\left[-\frac{1}{2}\right]$

$=\left[\frac{1}{2}\right]+\left[\frac{3}{2}\right]$

$=\left[\frac{4}{2}\right]$

=2 = R.H.S.
Hence Proved

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Question 5 E 

Prove the following :

sin90° = 2sin45°.cos45°
Sol :
We know that,
sin (90^o) = 1

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Taking LHS = sin 90° = 1

Now, taking RHS

$=2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

$=2 \times \frac{1}{2}$

= 1
= R.H.S.
Hence Proved

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Question 5 F 

Prove the following :

cos60° = 2cos²30° – 1 = 1 – 2 sin²30°
Sol :
We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Taking LHS = cos 60° $=\frac{1}{2}$

Now, solving RHS = 2cos² 30° - 1 , we get

$=2 \times\left(\frac{\sqrt{3}}{2}\right)^{2}-1$

$=2 \times \frac{3}{4}-1$

$=\frac{3}{2}-1$

$=\frac{3}{2}-1$

$=\frac{1}{2}$

= RHS

Now taking RHS = 1- 2sin² 30°

$=1-2\left(\frac{1}{2}\right)^{2}$

$=1-\frac{1}{2}$

$=\frac{2-1}{2}$

$=\frac{1}{2}$

= RHS
Hence, proved.

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Question 5 G 

Prove the following :

cos90° = 1 – 2 sin²45° = 2cos²45° – 1
Sol :
We know that
cos(90^o) = 0

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

taking LHS = cos 90° = 0

Now solving RHS 1- 2sin² 45°

$=1-2\left(\frac{1}{\sqrt{2}}\right)^{2}$

$=1-2 \times \frac{1}{2}$

= 1- 1
= 0
= RHS

Now, solving RHS = 2cos² 45° - 1 , we get

$=1-2 \times\left(\frac{1}{\sqrt{2}}\right)^{2}$

$=1-2 \times \frac{1}{2}$

= 1- 1
= 0
Hence, proved.

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Question 5 H 

Prove the following :

sin30°.cos60° + cos30°.sin60° = sin90°
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

sin (90^o) = 1

Taking LHS =
$=\left[\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\right]+\left[\left(\frac{\sqrt{3}}{2}\right) \times\left(\frac{\sqrt{3}}{2}\right)\right]$

$=\left[\left(\frac{1}{4}\right)\right]+\left[\left(\frac{3}{4}\right)\right]$

$=\left[\frac{1+3}{4}\right]$

= 1

Now, RHS = sin 90° = 1
∴ LHS = RHS
Hence, proved.

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Question 5 I 

Prove the following :

cos60°.cos30° – sin60°. sin30° = cos 90°
Sol :
We know that

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

cos(90^o) = 0

Taking LHS
$=\left[\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\right]-\left[\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)\right]$
$=\left[\left(\frac{\sqrt{3}}{4}\right)\right]-\left[\left(\frac{\sqrt{3}}{4}\right)\right]$
= 0

∴ LHS =RHS
Hence, proved.

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Question 5 J 

Prove the following :

$\cos 60^{\circ}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
Sol :
We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

Taking LHS = $\cos 60^{\circ}=\frac{1}{2}$

Now, solving RHS
$=\frac{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$
$=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}$
$=\frac{\frac{3-1}{3}}{\frac{3+1}{3}}$

$=\frac{2}{4}$

$=\frac{1}{2}$

∴ L.H.S. = R.H.S.
Hence, proved.

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Question 5 K 

Prove the following :

$\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \cdot \tan 30^{\circ}}=\tan 30^{\circ}$
Sol :
We know that
tan(60^o) = √3
$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

Taking LHS
$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+(\sqrt{3}) \times\left(\frac{1}{\sqrt{3}}\right)}$
$=\frac{\frac{3-1}{\sqrt{3}}}{1+1}$
$=\frac{\frac{2}{\sqrt{3}}}{2}$
$=\frac{1}{\sqrt{3}}$

Now, RHS $=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

∴L.H.S. = R.H.S.
Hence, proved.

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Question 5 L 

Prove the following :

$\frac{1-\tan 30^{\circ}}{1+\tan 30^{\circ}}=\frac{1-\sin 60^{\circ}}{\cos 60^{\circ}}$
Sol :

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Taking LHS
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Multiplying and Dividing, LHS by (√3- 1)
$=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$ [(a)² – (b)² = (a+b)(a-b)]
$=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$
$=\frac{3+1-2 \sqrt{3}}{3-1}$
$=\frac{4-2 \sqrt{3}}{2}$

Multiplying and Dividing, LHS by 2
= 2- √3

Now, RHS
$=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}$
$=\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}$
= 2- √3

∴ LHS = RHS
Hence, proved.

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Question 5 M 

Prove the following :

$\frac{\sin 60^{\circ}+\cos 30^{\circ}}{\sin 30^{\circ}+\cos 60^{\circ}+1}=\cos 30^{\circ}$
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Taking LHS
$=\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{\frac{1}{2}+\frac{1}{2}+1}$
$=\frac{2 \times \frac{\sqrt{3}}{2}}{\frac{1+1+2}{2}}$
$=\frac{\sqrt{3}}{2}$

Now, RHS= $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
∴ LHS =RHS

Hence Proved

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Question 5 N 

Prove the following :

$\sin 60^{\circ}=2 \sin 30^{\circ} \cdot \cos 30^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

Taking LHS $=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

Now, solving RHS = 2 sin 30° cos 30°

$=2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}}{2}$

= LHS

Now, RHS$=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
$=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$
$=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$=\frac{2}{\sqrt{3}} \times \frac{3}{4}$
$=\frac{\sqrt{3}}{2}$
∴ LHS =RHS
Hence proved

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Question 6A 

If A=60^o and B = 30^o, verify that :

cos (A+B) = cos A cos B – sin A sin B
Sol :
Given: A=60^o and B =30^o
Now, LHS = Cos (A+B)
⇒ Cos (60 ^o + 30 ^o)
⇒ Cos (90 ^o)
⇒ 0 [∵ cos 90 ^o = 0]
Now, RHS = Cos A Cos B – Sin A Sin B
⇒ cos(60 ^o) cos(30 ^o) – sin(60 ^o) sin (30 ^o)
$\Rightarrow\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
⇒ 0
∴ LHS = RHS
Hence Proved

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Question 6 B 

If A=60^o and B = 30^o, verify that :

sin (A – B) = sin A cos B – cos A sin B
Sol :
Given: A=60^o and B =30^o
Now, LHS = Sin (A-B)
⇒ Sin (60 ^o - 30 ^o)
⇒ Sin (30 ^o)
$\Rightarrow\left(\frac{1}{2}\right)$

Now, RHS = Sin A Cos B – Cos A Sin B

⇒ sin(60 ^o) cos(30 ^o) – cos(60 ^o) sin (30 ^o)
$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
$\Rightarrow \frac{3}{4}-\frac{1}{4}$
$\Rightarrow\left(\frac{1}{2}\right)$
∴ LHS = RHS
Hence Proved

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Question 6 C 

If A=60^o and B = 30^o, verify that :

tan (A – B) $=\frac{\tan A-\tan B}{1+\tan A \tan B}$
Sol :
Given: A=60^o and B =30^o
Now, LHS = tan (A-B)
⇒ tan (60 ^o - 30 ^o)
⇒ tan (30 ^o)
$\Rightarrow\left(\frac{1}{\sqrt{3}}\right)$

Now, RHS $=\frac{\tan A-\tan B}{1+\tan A \tan B}$
$=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$
$\Rightarrow \frac{\sqrt{3}-\left(\frac{1}{\sqrt{3}}\right)}{1+(\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)}$
$\Rightarrow \frac{\frac{3-1}{\sqrt{3}}}{1+1}$
$\Rightarrow\left(\frac{1}{\sqrt{3}}\right)$
∴ LHS = RHS
Hence Proved

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Question 7 A 

If A = 30^o, verify that :

sin 2A = 2 sin A cos A
Sol :
Given: A =30^o
Now, LHS = sin 2(30^o)
⇒ sin 60^o
$\Rightarrow \frac{\sqrt{3}}{2}$

Now, RHS = 2 sin A cos A
⇒ 2 sin (30^o) cos (30^o)
$\Rightarrow 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
$\Rightarrow \frac{\sqrt{3}}{2}$
∴ LHS = RHS
Hence Proved

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Question 7 B 

If A = 30^o, verify that :

cos 2A = 1-2 sin²A=2cos² A – 1
Sol :
Given: A =30^o

Now, LHS = cos 2(30^o)
⇒ cos 60^o
$\Rightarrow \frac{1}{2}$

Now, RHS = 1- 2sin² A
⇒ 1- 2sin² (30^o)
$\Rightarrow 1-2\left(\frac{1}{2}\right)^{2}$
$\Rightarrow 1-2\left(\frac{1}{4}\right)$
$\Rightarrow \frac{2-1}{2}$
$\Rightarrow \frac{1}{2}$

Now, RHS = 2cos² A – 1
⇒ 2cos² (30^o) - 1
$\Rightarrow 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1$
$\Rightarrow 2\left(\frac{3}{4}\right)-1$
$\Rightarrow \frac{3-2}{2}$
$\Rightarrow \frac{1}{2}$

∴ LHS = RHS
Hence Proved

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Question 8 A 

If θ = 30°, verify that :
sin 3θ = 3 sinθ – 4 sin³θ
Sol :
Given: θ =30^o
Now, LHS = sin 3(30^o)
⇒ sin 90^o
= 1
Now, RHS = 3 sin θ - 4 sin³ θ
⇒ 3 sin (30^o) - 4 sin³ (30^o)
$\Rightarrow 3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^{3}$
$\Rightarrow \frac{3}{2}-\frac{1}{2}$
= 1
∴ LHS = RHS
Hence Proved

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Question 8 B 

If θ = 30°, verify that :

cos3θ = 4cos³θ – 3cosθ
Sol :
Given: θ =30^o
Now, LHS = cos 3(30^o)
⇒ cos 90^o
= 0
Now, RHS = 4 cos³ θ - 3 cos θ
⇒ 4 cos³ (30^o) - 3 cos (30^o)
$\Rightarrow 4\left(\frac{\sqrt{3}}{2}\right)^{3}-3\left(\frac{\sqrt{3}}{2}\right)$
$\Rightarrow\left(\frac{3 \sqrt{3}}{2}\right)-\left(\frac{3 \sqrt{3}}{2}\right)$
= 0
∴ LHS = RHS
Hence Proved

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Question 9 

If sin (A + B) = 1 and cos (A – B) = $\frac{\sqrt{3}}{2}$, then find A and B.

Sol :
Given : sin (A+B) =1
⇒ Sin(A+B) = sin (90 ^o) [∵ sin (90 ^o)=1]
On equating both the sides, we get
A + B = 90 ^o …(1)
And $\cos (A-B)=\frac{\sqrt{3}}{2}$
⇒ cos(A – B) = cos (30 ^o) $\left[\because \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}\right]$

On equating both the sides, we get
A – B = 30 ^o …(2)

On Adding Eq. (1) and (2), we get
2A = 120 ^o
⇒ A = 60 ^o

Now, Putting the value of A in Eq.(1), we get
60 ^o + B =90 ^o
⇒ B = 30 ^o

Hence, A = 60 ^o and B = 30 ^o

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Question 10 

If sin (A + B) = 1 and cos (A – B) = 1, find A and B.

Sol :
Given : sin (A+B) =1
⇒ Sin(A+B) = sin (90 ^o) [∵ sin (90 ^o) =1]
On equating both the sides, we get
A + B = 90 ^o …(1)
And cos (A – B) = 1
⇒ cos(A – B) = cos (0 ^o) [∵ cos(0 ^o) = 1]
On equating both the sides, we get
A – B = 0 ^o …(2)
On Adding Eq. (1) and (2), we get
2A = 90 ^o
⇒ A = 45 ^o
Now, Putting the value of A in Eq.(1), we get
45 ^o + B =90 ^o
⇒ B = 45 ^o
Hence, A = 45 ^o and B = 45 ^o

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Question 11 

If sin (A + B) = cos (A – B) = $\frac{\sqrt{3}}{2}$, fins A and B.

Sol :
Given : $\sin (A+B)=\frac{\sqrt{3}}{2}$
⇒ Sin(A+B) = sin (60 ^o) $\left[\because \sin (60 ^{\circ})=\frac{\sqrt{3}}{2}\right]$

On equating both the sides, we get
A + B = 60 ^o …(1)
And $\cos (A-B)=\frac{\sqrt{3}}{2}$

⇒ cos(A – B) = cos (30 ^o) $\left[\because \cos (30^{\circ})=\frac{\sqrt{3}}{2}\right]$

On equating both the sides, we get
A – B = 30 ^o …(2)

On Adding Eq. (1) and (2), we get
2A = 90 ^o
⇒ A = 45 ^o

Now, Putting the value of A in Eq.(1), we get
45 ^o + B =60 ^o
⇒ B = 15 ^o
Hence, A = 45 ^o and B = 15 ^o

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Question 12 

If sin (A – B) = 1/2, cos(A + B) = 1/2; 0^o<A+B<90^o; A > B, find A and B.

Sol :
Given : $\sin (A-B)=\frac{1}{2}$
⇒ Sin(A-B) = sin (30 ^o) $\left[\because \sin (30^{\circ})=\frac{1}{2}\right]$

On equating both the sides, we get
A - B = 30 ^o …(1)
And $\cos (A+B)=\frac{1}{2}$

⇒ cos(A + B) = cos (60 ^o) $\left[\because \cos (60^{\circ})=\frac{1}{2}\right]$

On equating both the sides, we get
A + B = 60 ^o …(2)

On Adding Eq. (1) and (2), we get
2A = 90 ^o
⇒ A = 45 ^o

Now, Putting the value of A in Eq.(2), we get
45 ^o + B =60 ^o
⇒ B = 15 ^o
Hence, A = 45 ^o and B = 15 ^o

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Question 13 A 

Show by an example that

cos A – cos B ≠ cos (A – B)
Sol :

Let A = 60^o and B = 30^o, then
L.H.S. =cos A-cosB

=cos60°-cos30°

$=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}$

R. H. S=cos(A-B)

=cos(60°-30°)

$=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$

∴ L.H.S. R.H.S

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Question 13 B 

Show by an example that

cos C + cos D ≠ cos (C + D)
Sol :
Let C = 60^o and D = 30^o, then
L.H.S. = cos C + cos D = cos 60^o + cos 30^o
$=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}$
R. H. S. = cos (C+D) = cos (60^o + 30^o) = cos 90^o= 0
∴ L.H.S. R.H.S

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Question 13 C 

Show by an example that

sin A + sin B ≠ sin (A + B)
Sol :
Let A = 60^o and B = 30^o, then
L.H.S. = sin A + sin B = sin 60^o + sin 30^o
$=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}$

R. H. S. = sin (A + B) = sin (60^o + 30^o) = sin 90^o =1
∴ L.H.S. R.H.S

---

Question 13 D 

Show by an example that

sin A – sin B ≠ sin (A – B)
Sol :
Let A = 60^o and B = 30^o, then
L.H.S. = sin A - sin B = sin 60^o - sin 30^o
$=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}$

R. H. S. = sin (A - B) = sin (60^o - 30^o) = sin 30^o
$=\frac{1}{2}$
∴ L.H.S. ≠R.H.S

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Question 14 

In a right ΔABC hypotenuse AC = 10 cm and ∠A = 60°, then find the length of the remaining sides.

Sol :

Given: ∠A = 60^o and AC = 10cm
Now, $\sin 60^{\circ}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{BC}}{10}$

Now, we know that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{B C}{10}$
⇒ BC = 5√3 cm
In right angled ∆ABC , we have
⇒ (AB)² + (BC)² =(AC)² [by using Pythagoras theorem]
⇒ (AB)² + (5√3)² = (10)²
⇒ (AB)² +(25×3) =100
⇒ (AB)² +75 = 100
⇒ (AB)² = 100 – 75
⇒ (AB)² = 25
⇒ AB =√25
⇒ AB = ±5
⇒ AB = 5cm [taking positive square root since, side cannot be negative]
∴ Length of the side AB = 5cm and BC =5√3 cm

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Question 15 

In a rectangle ABCD, BD : BC = 2 : √3, then find ∠BDC in degrees.

Sol :

Given BD: BC = 2 : √3
We have to find the ∠BDC
We know that,
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\Rightarrow \sin \theta=\frac{k \sqrt{3}}{2 k}$
$\Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$
⇒ sin θ = sin 60^o
⇒ θ = 60^o

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