RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios

Class 10: Maths Chapter 10 solutions. Complete Class 10 Maths Chapter 10 Notes.

RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios

RS Aggarwal 10th Maths Chapter 1, Class 10 Maths Chapter 10 solutions

Page No 546:

Question 1:

If sin $\mathrm{\theta }=\frac{\sqrt{3}}{2}$, find the value of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that sin $\theta$ = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$= $\frac{AB}{AC}$ = $\frac{\sqrt{3}}{2 }$.

So, if AB = $\sqrt{3}k$, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² 
⇒ BC² = AC² $-$ AB² = (2k)² $-$ ($\sqrt{3}k$)²
⇒ BC² = 4k² $-$ 3k² = k²
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
   cos $\theta$  = $\frac{BC}{AC}$ = $\frac{k}{2k} = \frac{1}{2}$
   tan $\theta$  = $\frac{AB}{BC} = \frac{\sqrt{3}k}{k} = \sqrt{3}$

 ∴ cot $\theta$  = $\frac{1}{\tan \theta } = \frac{1}{\sqrt{3}}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \frac{2}{\sqrt{3}}$ and sec $\theta$  = $\frac{1}{\cos \theta } = 2$

Page No 546:

Question 2:

If cos $\mathrm{\theta }=\frac{7}{25}$ find the values of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cos $\theta$ = $\frac{\mathrm{Base}}{\mathrm{hypotenuse}}$= $\frac{BC}{AC}$ = $\frac{7}{25}$.

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² 
⇒ AB² = AC² $-$ BC² = (25k)² $-$ (7k)².
⇒ AB² = 625k² $-$ 49k² = 576k²
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
   sin $\theta$ = $\frac{AB}{AC}$ = $\frac{24k}{25k} = \frac{24}{25}$
   tan $\theta$ = $\frac{AB}{BC} = \frac{24k}{7k} = \frac{24}{7}$
 ∴ cot $\theta$ = $\frac{1}{\tan \theta } = \frac{7}{24}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \frac{25}{24}$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{25}{7}$

Page No 546:

Question 3:

If tan $\mathrm{\theta }=\frac{15}{8}$ find the values of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$ = $\frac{AB}{BC}$ = $\frac{15}{8}$.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (15k)² + (8k)²
⇒ AC² = 225k² + 64k² = 289k²
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
  sin $\theta$  = $\frac{AB}{AC}$ = $\frac{15k}{17k} = \frac{15}{17}$
  cos $\theta$  = $\frac{BC}{AC} = \frac{8k}{17k} = \frac{8}{17}$

∴ cot $\theta$  = $\frac{1}{\tan \theta } = \frac{8}{15}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \frac{17}{15}$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{17}{8}$

Page No 546:

Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cot $\theta$ = $\frac{\mathrm{base}}{\mathrm{Perpendicular}}$ = $\frac{BC}{AB}$ = 2.


So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (2k)² + (k)²
⇒ AC² = 4k² + k² = 5k²
⇒ AC = $\sqrt{5}$k
Now, finding the other T-ratios using their definitions, we get:
   sin $\theta$  = $\frac{AB}{AC}$ = $\frac{k}{\sqrt{5}k} = \frac{1}{\sqrt{5}}$
   cos $\theta$  = $\frac{BC}{AC} = \frac{2k}{\sqrt{5}k} = \frac{2}{\sqrt{5}}$

∴ tan $\theta$  = $\frac{1}{cot \theta } = \frac{1}{2}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \sqrt{5}$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{\sqrt{5}}{2}$

Page No 546:

Question 5:

If cosec θ = $\sqrt{10}$, the find the values of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cosec $\theta$ = $\frac{\mathrm{Hypotenuse}}{\mathrm{Perpendicular}}$ = $\frac{AC}{AB}$= $\frac{\sqrt{10}}{1}$.

So, if AC = ($\sqrt{10}$)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC² = AB² + BC² 
⇒ BC² = AC² $-$ AB² = 10k² $-$ k²
⇒ BC² = 9k²
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
   tan $\theta$  = $\frac{AB}{BC}$ = $\frac{k}{3k} = \frac{1}{3}$

   cos $\theta$  = $\frac{BC}{AC} = \frac{3k}{\sqrt{10}k} = \frac{3}{\sqrt{10}}$

 ∴ $\sin \theta =\frac{1}{\cos ec \theta }=\frac{1}{\sqrt{10}}$, cot $\theta$  = $\frac{1}{\tan \theta } = 3$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{\sqrt{10}}{3}$

Page No 546:

Question 6:

If $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$, find the values of all T-ratios of $\theta$.

Answer:

We have $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$,

As,

$$\begin{gathered} {\cos }^2\theta =1-{\sin }^2\theta \\ =1-{\left(\frac{a^2-b^2}{a^2+b^2}\right)}^2 \\ =\frac{1}{1}-\frac{{\left(a^2-b^2\right)}^2}{{\left(a^2+b^2\right)}^2} \\ =\frac{{\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2}{{\left(a^2+b^2\right)}^2} \\ =\frac{\left[\left(a^2+b^2\right)-\left(a^2-b^2\right)\right]\left[\left(a^2+b^2\right)+\left(a^2-b^2\right)\right]}{{\left(a^2+b^2\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{\left[a^2+b^2-a^2+b^2\right]\left[a^2+b^2+a^2-b^2\right]}{{\left(a^2+b^2\right)}^2} \\ =\frac{\left[2b^2\right]\left[2a^2\right]}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow {\cos }^2\theta =\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow \cos \theta =\sqrt{\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2}} \\ \Rightarrow \cos \theta =\frac{2ab}{\left(a^2+b^2\right)} \end{gathered}$$

Also,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{\left({\displaystyle \frac{a^2-b^2}{a^2+b^2}}\right)}{\left({\displaystyle \frac{2ab}{a^2+b^2}}\right)} \\ =\frac{a^2-b^2}{2ab} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{cosec}\theta =\frac{1}{\sin \theta } \\ =\frac{1}{\left({\displaystyle \frac{a^2-b^2}{a^2+b^2}}\right)} \\ =\frac{a^2+b^2}{a^2-b^2} \end{gathered}$$

Also,

$$\begin{gathered} \sec \theta =\frac{1}{\cos \theta } \\ =\frac{1}{\left({\displaystyle \frac{2ab}{a^2+b^2}}\right)} \\ =\frac{a^2+b^2}{2ab} \end{gathered}$$

And,

$$\begin{gathered} \cot \theta =\frac{1}{\tan \theta } \\ =\frac{1}{\left({\displaystyle \frac{a^2-b^2}{2ab}}\right)} \\ =\frac{2ab}{a^2-b^2} \end{gathered}$$

Page No 546:

Question 7:

If $\mathrm{sin\theta }=\frac{c}{\sqrt{c^2+d^2}}$, where d > 0 then find the values of cos θ and tan θ.

Answer:

$$\begin{gathered} \mathrm{Given}: \sin \theta =\frac{c}{\sqrt{c^2+d^2}} \\ \mathrm{Since}, \sin \theta =\frac{P}{H} \\ \Rightarrow P=c \mathrm{and} H=\sqrt{c^2+d^2} \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow c^2+B^2=c^2+d^2 \\ \Rightarrow B^2=d^2 \\ \Rightarrow B=d \\ \mathrm{Therefore}, \\ \cos \theta =\frac{B}{H}=\frac{d}{\sqrt{c^2+d^2}} \\ \tan \theta =\frac{P}{B}=\frac{c}{d} \\ \mathrm{Hence}, \cos \theta =\frac{d}{\sqrt{c^2+d^2}} \mathrm{and} \tan \theta =\frac{c}{d}. \end{gathered}$$

Page No 546:

Question 8:

If $\sqrt{3} \tan \mathrm{\theta }=1$ then evaluate (cos²θ – sin²θ).

Answer:

$$\begin{gathered} \mathrm{Given}: \sqrt{3}\tan \theta =1 \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \mathrm{Since}, \tan \theta =\frac{P}{B} \\ \Rightarrow P=1 \mathrm{and} B=\sqrt{3} \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+{\left(\sqrt{3}\right)}^2=H^2 \\ \Rightarrow H^2=1+3=4 \\ \Rightarrow H=2 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{1}{2} \\ \cos \theta =\frac{B}{H}=\frac{\sqrt{3}}{2} \\ {\cos }^2\theta -{\sin }^2\theta ={\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2 \\ =\frac{3}{4}-\frac{1}{4}=\frac{2}{4} \\ =\frac{1}{2} \\ \mathrm{Hence}, {\cos }^2\theta -{\sin }^2\theta =\frac{1}{2}. \end{gathered}$$

Page No 546:

Question 9:

If 4tan θ = 3 then prove that $\sin \mathrm{\theta } \cos \mathrm{\theta }=\frac{12}{25}$.

Answer:

$$\begin{gathered} \mathrm{Given}: 4\tan \theta =3 \\ \Rightarrow \tan \theta =\frac{3}{4} \\ \mathrm{Since}, \tan \theta =\frac{P}{B} \\ \Rightarrow P=3 \mathrm{and} B=4 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 3^2+4^2=H^2 \\ \Rightarrow H^2=9+16=25 \\ \Rightarrow H=5 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{3}{5} \\ \cos \theta =\frac{B}{H}=\frac{4}{5} \\ \sin \theta \times \cos \theta =\frac{3}{5}\times \frac{4}{5} \\ =\frac{12}{25} \\ \mathrm{Hence}, \sin \theta \times \cos \theta =\frac{12}{25}. \end{gathered}$$

Page No 546:

Question 10:

If $\sin \theta =\frac{a}{b}$, show that $\left(\sec \theta +\tan \theta \right)=\sqrt{\frac{b+a}{b-a}}$.

Answer:

$$\begin{gathered} \mathrm{LHS}=\left(\sec \theta +\tan \theta \right) \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\sqrt{1-{\sin }^2\theta }} \\ =\frac{\left(1+{\displaystyle \frac{a}{b}}\right)}{\sqrt{1-{\left({\displaystyle \frac{a}{b}}\right)}^2}} \end{gathered}$$
$$\begin{gathered} =\frac{\left({\displaystyle \frac{1}{1}+\frac{a}{b}}\right)}{\sqrt{{\displaystyle \frac{1}{1}-\frac{a^2}{b^2}}}} \\ =\frac{\left({\displaystyle \frac{b+a}{b}}\right)}{\sqrt{{\displaystyle \frac{b^2-a^2}{b^2}}}} \\ =\frac{\left({\displaystyle \frac{b+a}{b}}\right)}{\left({\displaystyle \frac{\sqrt{b^2-a^2}}{b}}\right)} \\ =\frac{\left({\displaystyle b+a}\right)}{{\displaystyle \sqrt{\left(b+a\right)\left(b-a\right)}}} \end{gathered}$$
$$\begin{gathered} =\frac{\left({\displaystyle b+a}\right)}{{\displaystyle \sqrt{\left(b+a\right)}\sqrt{\left(b-a\right)}}} \\ =\frac{\sqrt{\left(b+a\right)}}{{\displaystyle \sqrt{\left(b-a\right)}}} \\ =\sqrt{\frac{b+a}{b-a}} \\ =\mathrm{RHS} \end{gathered}$$

Page No 547:

Question 11:

If tan θ = $\frac{a}{b}$, show that $\left(\frac{a \mathrm{sin\theta }-b \mathrm{cos\theta }}{a \mathrm{sin\theta }+b \mathrm{cos\theta }}\right)=\frac{\left(a^2-b^2\right)}{\left(a^2+b^2\right)}.$

Answer:

It is given that tan $\theta = \frac{a}{b}$.

LHS = $\frac{a \sin \theta - b \cos \theta }{a \sin \theta + b \cos \theta }$
 Dividing the numerator and denominator by cos $\theta$, we get:

 $\frac{a \tan \theta - b}{a \tan \theta + b}$       (∵ tan $\theta = \frac{\sin \theta }{\cos \theta }$)
Now, substituting the value of tan $\theta$ in the above expression, we get:
 $$\begin{gathered} \frac{a\left({\displaystyle \frac{a}{b}}\right) - b}{a\left({\displaystyle \frac{a}{b}}\right) + b} \\ = \frac{{\displaystyle \frac{a^2}{b}} - b}{{\displaystyle \frac{a^2}{b}} + b} \\ = \frac{a^2 - b^2}{a^2 + b^2} = \mathrm{RHS} \end{gathered}$$
  i.e., LHS = RHS

 Hence proved.

Page No 547:

Question 12:

If sin θ=1213$\sin \theta =\frac{1213}{}$ then evaluate (2sin θ−3cos θ4sin θ−9cos θ)$\left(\frac{2\sin \theta -3\cos \theta 4\sin \theta -9\cos \theta }{}\right)$.

Answer:

$$\begin{gathered} \mathrm{Given}: \sin \theta =\frac{12}{13} \\ \mathrm{Since}, \sin \theta =\frac{P}{H} \\ \Rightarrow P=12 \mathrm{and} H=13 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow {12}^2+B^2={13}^2 \\ \Rightarrow B^2=169-144 \\ \Rightarrow B^2=25 \\ \Rightarrow B=5 \\ \mathrm{Therefore}, \\ \cos \theta =\frac{B}{H}=\frac{5}{13} \\ \mathrm{Now}, \\ \left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)=\left(\frac{2\left({\displaystyle \frac{12}{13}}\right)-3\left({\displaystyle \frac{5}{13}}\right)}{4\left({\displaystyle \frac{12}{13}}\right)-9\left({\displaystyle \frac{5}{13}}\right)}\right) \\ =\left(\frac{{\displaystyle \frac{24}{13}}-{\displaystyle \frac{15}{13}}}{{\displaystyle \frac{48}{13}}-{\displaystyle \frac{45}{13}}}\right) \\ =\left(\frac{{\displaystyle 24-15}}{{\displaystyle 48-45}}\right) \\ =\frac{{\displaystyle 9}}{{\displaystyle 3}} \\ =3 \\ \mathrm{Hence}, \left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)=3. \end{gathered}$$

Page No 547:

Question 13:

If $\tan \mathrm{\theta }=\frac{1}{2}$ then evaluate $\left(\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }}\right)$

Answer:

$$\begin{gathered} \mathrm{Given}: \tan \theta =\frac{1}{2} \\ \mathrm{Since}, \tan \theta =\frac{P}{B} \\ \Rightarrow P=1 \mathrm{and} B=2 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+2^2=H^2 \\ \Rightarrow H^2=1+4 \\ \Rightarrow H^2=5 \\ \Rightarrow H=\sqrt{5} \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{1}{\sqrt{5}} \\ \cos \theta =\frac{B}{H}=\frac{2}{\sqrt{5}} \\ \mathrm{Now}, \\ \left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{1+\cos \theta }\right)=\left(\frac{{\displaystyle \frac{2}{\sqrt{5}}}}{{\displaystyle \frac{1}{\sqrt{5}}}}+\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{1+{\displaystyle \frac{2}{\sqrt{5}}}}\right) \\ =\left(\frac{2}{1}+\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{{\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}}}}\right) \\ =\left(\frac{2}{1}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{5}+2}}\right) \\ =\left(2+\left(\frac{{\displaystyle 1}}{{\displaystyle \sqrt{5}+2}}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}\right)\right) \\ =\left(2+\left(\frac{\sqrt{5}-2}{5-4}\right)\right) \\ =\left(2+\sqrt{5}-2\right) \\ =\sqrt{5} \\ \mathrm{Hence}, \left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{1+\cos \theta }\right)=\sqrt{5}. \end{gathered}$$

Page No 547:

Question 14:

If sinα=12$\mathrm{sin\alpha }=\frac{12}{}$, prove that (3cosα−4cos3α)=0$\left(3\mathrm{cos\alpha }-4{\cos }^3\alpha \right)=0$.

Answer:

$$\begin{gathered} \mathrm{LHS}=\left(3\cos \alpha -4{\cos }^3\alpha \right) \\ =\cos \alpha \left(3-4{\cos }^2\alpha \right) \\ =\sqrt{1-{\sin }^2\alpha }\left[3-4\left(1-{\sin }^2\alpha \right)\right] \\ =\sqrt{1-{\left(\frac{1}{2}\right)}^2}\left[3-4\left(1-{\left(\frac{1}{2}\right)}^2\right)\right] \\ =\sqrt{\frac{1}{1}-\frac{1}{4}}\left[3-4\left(\frac{1}{1}-\frac{1}{4}\right)\right] \\ =\sqrt{\frac{3}{4}}\left[3-4\left(\frac{3}{4}\right)\right] \\ =\sqrt{\frac{3}{4}}\left[3-3\right] \\ =\sqrt{\frac{3}{4}}\left[0\right] \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

Page No 547:

Question 15:

If 3 cot θ = 2, show that $\left(\frac{4\mathrm{sin\theta }-3\mathrm{cos\theta }}{2\mathrm{sin\theta }+6\mathrm{cos\theta }}\right)=\frac{1}{3}.$

Answer:

It is given that cot $\theta = \frac{2}{3}$.

LHS  = $\frac{4 \sin \theta - 3 \cos \theta }{2 \sin \theta + 6 \cos \theta }$
Dividing the above expression by sin $\theta$, we get:
$\frac{4 - 3 cot \theta }{2 + 6 cot \theta }$                     [∵ cot $\theta = \frac{\cos \theta }{\sin \theta }$]
Now, substituting the values of cot $\theta$ in the above expression, we get:
$$\begin{gathered} \frac{4 - 3\left({\displaystyle \frac{2}{3}}\right)}{2 + 6\left({\displaystyle \frac{2}{3}}\right)} \\ = \frac{4 - 2}{2 + 4} = \frac{2}{6}=\frac{1}{3} \end{gathered}$$
 i.e., LHS = RHS
 
Hence proved.

Page No 547:

Question 16:

If sec θ = $\frac{17}{8}$ then prove that $\frac{3-4{\sin }^2\mathrm{\theta }}{4{\cos }^2\mathrm{\theta }-3}=\frac{3-{\tan }^2\mathrm{\theta }}{1-3{\tan }^2\mathrm{\theta }}$.

Answer:

It is given that sec $\theta$ = $\frac{17}{8}$.

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that cos $\theta$ = $\frac{1}{sec \theta }= \frac{8}{17} = \frac{BC}{AC}$
 
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (17k)² $-$ (8k)²
⇒ AB² = 289k² $-$ 64k² = 225k²
⇒ AB = 15k.

Now, tan $\theta$  = $\frac{AB}{BC} = \frac{15}{8}$ and sin $\theta$ = $\frac{AB}{AC} = \frac{15k}{17k}= \frac{15}{17}$

The given expression is $\frac{3 - 4{\sin }^2\theta }{4{\cos }^2\theta - 3} = \frac{3 - {\tan }^2\theta }{1 - 3{\tan }^2\theta }$.
 
 Substituting the values in the above expression, we get:
 $$\begin{gathered} \mathrm{LHS}= \frac{3 - 4{\left({\displaystyle \frac{15}{17}}\right)}^2}{4{\left({\displaystyle \frac{8}{17}}\right)}^2 - 3} \\ = \frac{3 - {\displaystyle \frac{900}{289}}}{{\displaystyle \frac{256}{289}}- 3} \\ = \frac{867-900}{256-867}= \frac{-33}{-611}=\frac{33}{611} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS} = \frac{3-{\left({\displaystyle \frac{15}{8}}\right)}^2}{1-3{\left({\displaystyle \frac{15}{8}}\right)}^2} \\ =\frac{3-{\displaystyle \frac{225}{64}}}{1-{\displaystyle \frac{675}{64}}} \\ =\frac{192-225}{64-675}=\frac{-33}{-611}=\frac{33}{611} \end{gathered}$$

∴ LHS = RHS
Hence proved.

Page No 547:

Question 17:

If tan θ = $\frac{20}{21}$, show that$\frac{\left(1-\mathrm{sin\theta }+\mathrm{cos\theta }\right)}{\left(1+\mathrm{sin\theta }+\mathrm{cos\theta }\right)}=\frac{3}{7}.$

Answer:

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{AB}{BC}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
 AC² = AB² + BC²
⇒ AC²= (20k)² + (21k)²
⇒ AC² = 841k²
⇒  AC = 29k
Now, sin $\theta$ = $\frac{AB}{AC} = \frac{20}{29}$ and cos $\theta$ = $\frac{BC}{AC} = \frac{21}{29}$

Substituting these values in the given expression, we get:
 $$\begin{gathered} \mathrm{LHS}=\frac{1 - \sin \theta + \cos \theta }{1 + \sin \theta + \cos \theta } \\ = \frac{1 - {\displaystyle \frac{20}{29}} + {\displaystyle \frac{21}{29}}}{1 + {\displaystyle \frac{20}{29}} + {\displaystyle \frac{21}{29}}} \\ = \frac{{\displaystyle \frac{29 - 20 + 21}{29}}}{{\displaystyle \frac{29 + 20 + 21}{29}}}= \frac{30}{70} = \frac{3}{7} = \mathrm{RHS} \end{gathered}$$
∴ LHS = RHS

Hence proved.

Page No 547:

Question 18:

If tan θ = $\frac{1}{\sqrt{7}}$ then prove that $\left(\frac{{\mathrm{cosec}}^2\mathrm{\theta }+{\sec }^2\mathrm{\theta }}{{\mathrm{cosec}}^2\mathrm{\theta }-{\sec }^2\mathrm{\theta }}\right)=\frac{4}{3}.$

Answer:

Let us consider a right △△ABC, right-angled at B and ∠C=θ$\angle C=\theta$.
Now it is given that tan θ$\theta$ = ABBC$\frac{\mathrm{ABBC}}{}$ =  17√$\frac{1}{\sqrt{7}}$.

So, if AB = k, then BC = 7–√$\sqrt{7}$k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AC² = (k)² + (7–√$\sqrt{7}$k)²
⇒ AC² = k² + 7k²
⇒ AC = 22–√$\sqrt{2}$k
Now, finding out the values of the other trigonometric ratios, we have:
sin θ$\theta$  = ABAC = k22√k = 122√$\frac{\mathrm{ABAC}}{} = \frac{k}{2\sqrt{2}k} = \frac{1}{2\sqrt{2}}$
cos θ$\theta$  = BCAC = 7√ k22√k = 7√22√$\frac{\mathrm{BCAC}}{} = \frac{\sqrt{7} k2\sqrt{2}k}{} = \frac{\sqrt{7}}{2\sqrt{2}}$
∴ cosec θ$\theta$  = 1sin θ = 22–√$\frac{1}{\sin \theta } = 2\sqrt{2}$ and sec θ$\theta$   = 1cos θ = 22√7√$\frac{1}{\cos \theta } = \frac{2\sqrt{2}}{\sqrt{7}}$
Substituting the values of cosec θ$\theta$  and sec θ$\theta$  in the given expression, we get:
 cosec2θ − sec2θcosec2θ + sec2θ=(22√)2 − (22√7√)2(22√)2 + (22√7√)2=8 − (87)8 + (87)=56 − 8756 + 87=4864 = 34 = RHS$$\begin{gathered} \frac{\mathrm{cose}{c}^2\theta - \mathrm{se}{c}^2\mathrm{\theta cose}{c}^2\theta + \mathrm{se}{c}^2\theta }{}\text{ \\ =(22)2 - 2272(22)2 + 2272 \\ =8 - 878 + 87 \\ =56 - 8756 + 87 \\ =4864 = 34 = RHS} \end{gathered}$$
 i.e., LHS = RHS
 
Hence proved.

Page No 547:

Question 19:

If $\sin \theta =\frac{3}{4}$, show that $\sqrt{\frac{{\mathrm{cosec}}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\frac{\sqrt{7}}{3}$.

Answer:

$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{{\mathrm{cosec}}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}} \\ =\sqrt{\frac{1}{{\tan }^2\theta }} \\ =\sqrt{{\cot }^2\theta } \\ =\cot \theta \\ =\sqrt{{\mathrm{cosec}}^2\theta -1} \\ =\sqrt{{\left(\frac{1}{\sin \theta }\right)}^2-1} \\ =\sqrt{{\left(\frac{1}{\left({\displaystyle \frac{3}{4}}\right)}\right)}^2-1} \\ =\sqrt{{\left(\frac{4}{3}\right)}^2-1} \\ =\sqrt{\frac{16}{9}-1} \\ =\sqrt{\frac{16-9}{9}} \\ =\sqrt{\frac{7}{9}} \\ =\frac{\sqrt{7}}{3} \\ =\mathrm{RHS} \end{gathered}$$

Page No 547:

Question 20:

If 3 tan A = 4 then prove that
(i) $\sqrt{\frac{\sec A-\mathrm{cosec} A}{\sec A+\mathrm{cosec} A}}=\frac{1}{\sqrt{7}}$
(ii) $\sqrt{\frac{1-\sin A}{1+\cos A}}=\frac{1}{2\sqrt{2}}$

Answer:

(i)
 ​$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{\sec \theta -\mathrm{cosec}\theta }{\sec \theta +\mathrm{cosec}\theta }} \\ =\sqrt{\frac{\left({\displaystyle \frac{1}{\cos \theta }}-{\displaystyle \frac{1}{\sin \theta }}\right)}{\left({\displaystyle \frac{1}{\cos \theta }}+{\displaystyle \frac{1}{\sin \theta }}\right)}} \\ =\sqrt{\frac{\left({\displaystyle \frac{\sin \theta -\cos \theta }{\sin \theta \cos \theta }}\right)}{\left({\displaystyle \frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }}\right)}} \\ =\sqrt{\frac{\left({\displaystyle \frac{\sin \theta -\cos \theta }{\sin \theta }}\right)}{\left({\displaystyle \frac{\sin \theta +\cos \theta }{\sin \theta }}\right)}} \\ =\sqrt{\frac{\left({\displaystyle \frac{\sin \theta }{\sin \theta }-\frac{{\displaystyle \cos \theta }}{\sin \theta }}\right)}{\left({\displaystyle \frac{\sin \theta }{\sin \theta }}+{\displaystyle \frac{\cos \theta }{\sin \theta }}\right)}} \\ =\sqrt{\frac{1-\cot \theta }{1+\cot \theta }} \\ =\sqrt{\frac{\left(1-{\displaystyle \frac{3}{4}}\right)}{\left(1+{\displaystyle \frac{3}{4}}\right)}} \end{gathered}$$
$$\begin{gathered} =\sqrt{\frac{\left({\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{7}{4}}\right)}} \\ =\sqrt{\frac{1}{7}} \\ =\frac{1}{\sqrt{7}} \\ =\mathrm{RHS} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Given}: 3\tan A=4 \\ \Rightarrow \tan A=\frac{4}{3} \\ \mathrm{Since}, \tan A=\frac{P}{B} \\ \Rightarrow P=4 \mathrm{and} B=3 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 4^2+3^2=H^2 \\ \Rightarrow H^2=16+9 \\ \Rightarrow H^2=25 \\ \Rightarrow H=5 \\ \mathrm{Therefore}, \\ \sin A=\frac{P}{H}=\frac{4}{5} \\ \cos A=\frac{B}{H}=\frac{3}{5} \\ \mathrm{Now}, \\ \sqrt{\frac{1-\sin A}{1+\cos A}}=\sqrt{\frac{{\displaystyle 1-\frac{4}{5}}}{1+{\displaystyle \frac{3}{5}}}} \\ =\sqrt{\frac{{\displaystyle \frac{5-4}{5}}}{{\displaystyle \frac{5+3}{5}}}} \\ =\sqrt{\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{8}{5}}}} \\ =\sqrt{\frac{{\displaystyle 1}}{{\displaystyle 8}}} \\ =\frac{1}{2\sqrt{2}} \\ \mathrm{Hence}, \sqrt{\frac{1-\sin A}{1+\cos A}}=\frac{1}{2\sqrt{2}}. \end{gathered}$$

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Question 21:

If $\cot \mathrm{\theta }=\frac{15}{8} \mathrm{then} \mathrm{evaluate} \frac{\left(1+\sin \mathrm{\theta }\right) \left(1-\sin \mathrm{\theta }\right)}{\left(1+\cos \mathrm{\theta }\right) \left(1-\cos \mathrm{\theta }\right)}$.

Answer:

$$\begin{gathered} \mathrm{Given}: \cot \theta =\frac{15}{8} \\ \mathrm{Since}, \cot \theta =\frac{B}{P} \\ \Rightarrow P=8 \mathrm{and} B=15 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 8^2+{15}^2=H^2 \\ \Rightarrow H^2=64+225 \\ \Rightarrow H^2=289 \\ \Rightarrow H=17 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{8}{17} \\ \cos \theta =\frac{B}{H}=\frac{15}{17} \\ \mathrm{Now}, \\ \frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta } \\ =\frac{{\cos }^2\theta }{{\sin }^2\theta } \left(\because {\sin }^2\theta +{\cos }^2\theta =1\right) \\ ={\cot }^2\theta \\ ={\left(\frac{{\displaystyle 15}}{{\displaystyle 8}}\right)}^2 \\ =\frac{225}{64} \\ \mathrm{Hence}, \frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}=\frac{225}{64}. \end{gathered}$$

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Question 22:

In a right $∆\mathrm{ABC}$, right-angled at $\mathrm{B}$, if $\mathrm{tanA}=1$, then verify that $2\mathrm{sinA}·\mathrm{cosA}=1$.

Answer:

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{tanA}=1 \\ \Rightarrow \frac{\mathrm{sinA}}{\mathrm{cosA}}=1 \\ \Rightarrow \mathrm{sinA}=\mathrm{cosA} \\ \Rightarrow \mathrm{sinA}-\mathrm{cosA}=0 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ {\left(\mathrm{sinA}-\mathrm{cosA}\right)}^2=0 \\ \Rightarrow {\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}-2\mathrm{sinA}·\mathrm{cosA}=0 \\ \Rightarrow 1-2\mathrm{sinA}·\mathrm{cosA}=0 \\ \therefore 2\mathrm{sinA}·\mathrm{cosA}=1 \end{gathered}$$

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Question 23:

In the given figure, ABCD is a rectangle in which diag. AC = 17 cm, ∠BCA = θ and $\sin \mathrm{\theta }=\frac{8}{17}$.
Find (i) the area of rect. ABCD, (ii) the perimeter of rect. ABCD.

Answer:

$$\begin{gathered} \mathrm{Given}: \mathrm{In} ∆ABC, \\ AC=17 \mathrm{cm} \\ \sin \theta =\frac{8}{17} \\ \mathrm{Since}, \sin \theta =\frac{P}{H} \\ \Rightarrow P=8 \mathrm{and} H=17 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 8^2+B^2={17}^2 \\ \Rightarrow B^2=289-64 \\ \Rightarrow B^2=225 \\ \Rightarrow B=15 \\ \mathrm{Therefore}, \\ AB= 8 \mathrm{cm} \mathrm{and} BC=15 \mathrm{cm} \\ \mathrm{Therefore}, \\ \left(\mathrm{i}\right) \mathrm{Area} \mathrm{of} \mathrm{rectangle} ABCD=AB\times BC \\ =8\times 15 \\ =120 {\mathrm{cm}}^2 \\ \left(ii\right) \mathrm{Perimeter} \mathrm{of} \mathrm{rectangle} ABCD=2\left(AB+BC\right) \\ =2\left(8+15\right) \\ =2\left(23\right) \\ =46 \mathrm{cm} \end{gathered}$$

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Question 24:

If $x=\mathrm{cosecA}+\mathrm{cosA}$ and $y=\mathrm{cosecA}-\mathrm{cosA}$, then prove that ${\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2-1=0$.

Answer:

$$\begin{gathered} \mathrm{LHS}={\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2-1 \\ ={\left[\frac{2}{\left(\mathrm{cosecA}+\mathrm{cosA}\right)+\left(\mathrm{cosecA}-\mathrm{cosA}\right)}\right]}^2+{\left[\frac{\left(\mathrm{cosecA}+\mathrm{cosA}\right)-\left(\mathrm{cosecA}-\mathrm{cosA}\right)}{2}\right]}^2-1 \\ ={\left[\frac{2}{\mathrm{cosecA}+\mathrm{cosA}+\mathrm{cosecA}-\mathrm{cosA}}\right]}^2+{\left[\frac{\mathrm{cosecA}+\mathrm{cosA}-\mathrm{cosecA}+\mathrm{cosA}}{2}\right]}^2-1 \\ ={\left[\frac{2}{2\mathrm{cosecA}}\right]}^2+{\left[\frac{2\mathrm{cosA}}{2}\right]}^2-1 \end{gathered}$$
$$\begin{gathered} ={\left[\frac{1}{\mathrm{cosecA}}\right]}^2+{\left[\mathrm{cosA}\right]}^2-1 \\ ={\left[\mathrm{sinA}\right]}^2+{\left[\mathrm{cosA}\right]}^2-1 \\ ={\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}-1 \\ =1-1 \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

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Question 25:

If $x=\mathrm{cotA}+\mathrm{cosA}$ and $y=\mathrm{cotA}-\mathrm{cosA}$, prove that ${\left(\frac{x-y}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2=1$.

Answer:

$$\begin{gathered} \mathrm{LHS}={\left(\frac{x-y}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2 \\ ={\left[\frac{\left(\mathrm{cotA}+\mathrm{cosA}\right)-\left(\mathrm{cotA}-\mathrm{cosA}\right)}{\left(\mathrm{cotA}+\mathrm{cosA}\right)+\left(\mathrm{cotA}-\mathrm{cosA}\right)}\right]}^2+{\left[\frac{\left(\mathrm{cotA}+\mathrm{cosA}\right)-\left(\mathrm{cotA}-\mathrm{cosA}\right)}{2}\right]}^2 \\ ={\left[\frac{\mathrm{cotA}+\mathrm{cosA}-\mathrm{cotA}+\mathrm{cosA}}{\mathrm{cotA}+\mathrm{cosA}+\mathrm{cotA}-\mathrm{cosA}}\right]}^2+{\left[\frac{\mathrm{cotA}+\mathrm{cosA}-\mathrm{cotA}+\mathrm{cosA}}{2}\right]}^2 \\ ={\left[\frac{2\mathrm{cosA}}{2\mathrm{cotA}}\right]}^2+{\left[\frac{2\mathrm{cosA}}{2}\right]}^2 \\ ={\left[\frac{\mathrm{cosA}}{\left({\displaystyle \frac{\mathrm{cosA}}{\mathrm{sinA}}}\right)}\right]}^2+{\left[\mathrm{cosA}\right]}^2 \\ ={\left[\frac{\mathrm{sinA} \mathrm{cosA}}{\mathrm{cosA}}\right]}^2+{\left[\mathrm{cosA}\right]}^2 \\ ={\left[\mathrm{sinA}\right]}^2+{\left[\mathrm{cosA}\right]}^2 \\ ={\sin }^2\mathrm{A}+{\cos }^2\mathrm{A} \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

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Question 26:

In the figure of $∆\mathrm{PQR}, \angle \mathrm{P}=\theta ° \mathrm{and} \angle \mathrm{R}=\varphi °$. Find
$$\begin{gathered} \left(\mathrm{i}\right) \left(\sqrt{x+1}\right)\cot \varphi \\ \left(\mathrm{ii}\right) \left(\sqrt{x^3+x^2}\right)\tan \theta \\ \left(\mathrm{iii}\right) \cos \theta \end{gathered}$$

Answer:

In $∆\mathrm{PQR}, \angle \mathrm{Q}=90°$,

Using Pythagoras theorem, we get

$$\begin{gathered} \mathrm{PQ}=\sqrt{{\mathrm{PR}}^2-{\mathrm{QR}}^2} \\ =\sqrt{{\left(x+2\right)}^2-x^2} \\ =\sqrt{x^2+4x+4-x^2} \\ =\sqrt{4\left(x+1\right)} \\ =2\sqrt{x+1} \end{gathered}$$

Now,

$$\begin{gathered} \left(\mathrm{i}\right) \left(\sqrt{x+1}\right)\cot \varphi \\ =\left(\sqrt{x+1}\right)\times \frac{\mathrm{QR}}{\mathrm{PQ}} \\ =\left(\sqrt{x+1}\right)\times \frac{x}{2\sqrt{x+1}} \\ =\frac{x}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \left(\sqrt{x^3+x^2}\right)\tan \theta \\ =\left(\sqrt{x^2\left(x+1\right)}\right)\times \frac{\mathrm{QR}}{\mathrm{PQ}} \\ =x\sqrt{\left(x+1\right)}\times \frac{x}{2\sqrt{x+1}} \\ =\frac{x^2}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \cos \theta \\ =\frac{\mathrm{PQ}}{\mathrm{PR}} \\ =\frac{2\sqrt{x+1}}{\left(x+2\right)} \end{gathered}$$

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Question 27:

If $\cot A+\frac{1}{\cot A}=2$, find the value of $\left({\cot }^{2}A+\frac{1}{{\cot }^{2}A}\right)$.

Answer:

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}\cot A+\frac{1}{\cot A}=2 \\ \cot A+\frac{1}{\cot A}=2 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(\cot A+\frac{1}{\cot A}\right)}^2=2^2 \\ \Rightarrow {\cot }^{2}A+{\left(\frac{1}{\cot A}\right)}^2+2\left(\cot A\right)\left(\frac{1}{\cot A}\right)=4 \\ \Rightarrow {\cot }^{2}A+\frac{1}{{\cot }^{2}A}+2=4 \\ \Rightarrow {\cot }^{2}A+\frac{1}{{\cot }^{2}A}=4-2 \\ \Rightarrow {\cot }^{2}A+\frac{1}{{\cot }^{2}A}=2 \\ \mathrm{Hence}, \mathrm{the} \mathrm{value} \mathrm{of} \left({\cot }^{2}A+\frac{1}{{\cot }^{2}A}\right) \mathrm{is} 2. \end{gathered}$$

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RS Aggarwal Solutions for Class 10 Maths Chapter 10: Download PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios

Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios PDF

About Rs Aggarwal

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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