Exercise 3.3 (Revised) - Chapter 3 - Trigonometric Functions - Ncert Solutions class 11 - Maths

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Chapter 3: Trigonometric Functions | NCERT Solutions for Class 11 Maths

Prove that:
Ex 3.3 Question 1.

$\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}=-\frac{1}{2}$

Answer.

Taking L.H.S.
$
\begin{aligned}
& =\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4} \\
& =\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-(1)^2=\frac{1}{4}+\frac{1}{4}-1 \\
& =\frac{1+1-4}{4}=\frac{-2}{4}=\frac{-1}{2} \text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 2.

$2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2}$

Answer.

Taking L.H.S
$
\begin{aligned}
& =2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3} \\
& =2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2\left(\pi+\frac{\pi}{6}\right) \cos ^2 \frac{\pi}{3}
\end{aligned}
$

$=2 \sin ^2 \frac{\pi}{6}-\operatorname{cosec} 2 \frac{\pi}{6} \cos ^2 \frac{\pi}{3}$

$
\begin{aligned}
& =2 \times\left(\frac{1}{2}\right)^2+(-2)^2 \times\left(\frac{1}{2}\right)^2 \\
& =2 \times \frac{1}{4}+4 \times \frac{1}{4}=\frac{6}{4}=\frac{3}{2}
\end{aligned}
$
Ex 3.3 Question 3.

$\cot ^2 \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}=6$

Answer.

Taking L.H.S
$
\begin{aligned}
& =\cot ^2 \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6} \\
& =\cot ^2 \frac{\pi}{6}+\operatorname{cosec}\left(\pi-\frac{\pi}{6}\right)+3 \tan ^2 \frac{\pi}{6} \\
& =\cot ^2 \frac{\pi}{6}+\operatorname{cosec} \frac{\pi}{6}+3 \tan ^2 \frac{\pi}{6} \\
& =(\sqrt{3})^2+2+3 \times\left(\frac{1}{\sqrt{3}}\right)^2 \\
& =3+2+3 \times \frac{1}{3}=5+1=6=\text { R.H.S. }
\end{aligned}
$

Ex 3.3 Question 4.

$\text {} 2 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3}=10$

Answer:

$\begin{aligned}
& \text {L.H.S. }=2 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3} \\
& =2 \sin ^2\left(\pi-\frac{\pi}{4}\right)+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3} \\
& =2 \sin ^2 \frac{\pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3}
\end{aligned}$

$\begin{aligned}
& =2 \times\left(\frac{1}{\sqrt{2}}\right)^2+2 \times\left(\frac{1}{\sqrt{2}}\right)^2+2 \times(2)^2 \\
& =2 \times \frac{1}{2}+2 \times \frac{1}{2}+2 \times 4=1+1+8=10=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question 5.

Find the value of:
(i) $\sin 75^{\circ}$
(ii) $\tan 15^{\circ}$

Answer.

(i) $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}$
$
=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}
$
$
\begin{aligned}
& \text { (ii) } \tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} \\
& {\left[\because \tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right]} \\
& =\frac{1-\frac{1}{\sqrt{3}}}{1+1 \times \frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\
& =\frac{3+1-2 \sqrt{3}}{3-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}
\end{aligned}
$

Prove the following:
Ex 3.3 Question 6.

$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$

Answer.

Taking L.H.S
$
\begin{aligned}
& =\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right) \\
& =\cos \left[\frac{\pi}{4}-x+\frac{\pi}{4}-y\right] \\
& {[\because \cos (x+y)=\cos x \cos y-\sin x \sin y]} \\
& =\cos \left[\frac{\pi}{2}-(x+y)\right]=\sin (x+y)=\text { R.H.S. }
\end{aligned}
$

Ex 3.3 Question 7.

$\text {} \frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left[\frac{1+\tan x}{1-\tan x}\right]^2$

Answer.

Taking L.H.S
$
=\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}
$

$\text { [Using } \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}, \tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y} \text { ] }$

$
=\frac{(1+\tan x)^2}{(1-\tan x)^2}=\text { R.H.S. }
$
Ex 3.3 Question 8.

$\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^2 x$

Answer.

Taking L.H.S
$
\begin{aligned}
& =\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)} \\
& =\frac{-\cos x \cos x}{\sin x(-\sin x)}=\frac{-\cos ^2 x}{-\sin ^2 x}=\cot ^2 x=\text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 9.

$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$

Answer.

Taking L.H.S
$
=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]
$

$\begin{aligned}
& =\sin x \cos x(\tan x+\cot x) \\
& =\sin x \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right) \\
& =\sin x \cos x\left(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\right) \\
& =1=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question 10.

$\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x=\cos x$

Answer.

Taking L.H.S.
$
\begin{aligned}
& =\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x \\
& =\cos [(n+1) x-(n+2) x] \\
& =\cos [n x+x-n x-2 x] \\
& =\cos (-x)=\cos x=\text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 11.

$\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$

Answer.

Taking L.H.S
$
\begin{aligned}
& =\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right) \\
& =-2 \sin \frac{3 \pi}{4} \sin x=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x \\
& =-2 \sin \frac{\pi}{4} \sin x=-2 \times \frac{1}{\sqrt{2}} \sin x \\
& =-\sqrt{2} \sin x=\text { R.H.S. }
\end{aligned}
$

Ex 3.3 Question 12.

$\text {} \sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x$

Answer:

$\begin{aligned}
& \text {L.H.S. }=\sin ^2 6 x-\sin ^2 4 x \\
& =\sin (6 x+4 x) \cdot \sin (6 x-4 x) \\
& {\left[\because \sin ^2 x-\sin ^2 y=\sin (x+y) \sin (x-y)\right]} \\
& =\sin 10 x \sin 4 x=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question 13.

$\text {} \cos ^2 2 x-\cos ^2 6 x=\sin 4 x \sin 8 x$

Answer:

$\begin{aligned}
& \text {L.H.S. }=\cos ^2 2 x-\cos ^2 6 x \\
& =\sin (2 x+6 x) \cdot \sin (6 x-2 x) \\
& {\left[\because \cos ^2 y-\cos ^2 x=\sin (x+y) \sin (x-y)\right]} \\
& =\sin 8 x \sin 4 x=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question 14.

$\text {} \sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^2 x \sin 4 x$

Answer:

$\begin{aligned}
& \text {L.H.S. }=\sin 2 x+2 \sin 4 x+\sin 6 x \\
& =[\sin 4 x+\sin 2 x]+[\sin 6 x+\sin 4 x] \\
& =2 \sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right) \\
& =2 \sin 3 \mathrm{x} \cdot \cos \mathrm{x}+2 \sin 5 \mathrm{x} \cos \mathrm{x} \\
& =2 \cos x[\sin 3 x+\sin 5 x] \\
& =2 \cos x\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right] \\
& =2 \cos x[2 \sin 4 x \cos x] \\
& =4 \cos 2 x \sin 4 x=\text { R.H.S. }
\end{aligned}$
Ex 3.3 Question 15.

$\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$

Answer:

$
\text {L.H.S. }=\cot 4 x(\sin 5 x+\sin 3 x)
$

$=\frac{\cos 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]$

$
\begin{aligned}
& =\frac{\cos 4 x}{\sin 4 x}[2 \sin 4 x \cos x]=2 \cos 4 x \cos x \\
& \text { R.H.S. }=\cot \mathrm{x}(\sin 5 \mathrm{x}-\sin 3 \mathrm{x}) \\
& =\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right] \\
& =\frac{\cos x}{\sin x}[2 \cos 4 x \sin x] \\
& =2 \cos 4 x \cos x \\
& \therefore \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 16.

$\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$

Answer.

L.H.S. $=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}$
$
\begin{aligned}
& =\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 x+3 x}{2}\right) \sin \left(\frac{17 x-3 x}{2}\right)} \\
& =\frac{-2 \sin 7 x \sin 2 x}{2 \cos 10 x \sin 7 x}=-\frac{\sin 2 x}{\cos 10 x}=\text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 17.

$\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$

Answer:

$\begin{aligned}
& \text {L.H.S. }=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x} \\
& =\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)} \\
& =\frac{2 \sin 4 x}{2 \cos 4 x}=\tan 4 x=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question 18.

$\text {} \frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right)$

Answer:

$\begin{aligned}
& \text { L.H.S. }=\frac{\sin x-\sin y}{\cos x+\cos y} \\
& =\frac{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} \\
& =\frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)}=\tan \left(\frac{x-y}{2}\right)=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question 19.

$\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$

Answer.

L.H.S. $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$
$=\frac{\sin 2 x}{\cos 2 x}=\tan 2 x=$ R.H.S .
Ex 3.3 Question 20.

$\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}=2 \sin x$

Answer.

L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}$

$\begin{aligned}
& =\frac{-(\sin 3 x-\sin x)}{-\left(\cos ^2 x-\sin ^2 x\right)} \\
& =\frac{2 \cos \left(\frac{3 x+x}{2}\right) \sin \left(\frac{3 x-x}{2}\right)}{\cos 2 x} \\
& =\frac{2 \cos 2 x \sin x}{\cos 2 x}=2 \sin x=\text { R.H.S. }
\end{aligned}$

Ex 3.3 Question21.

$\text {} \frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$

Answer:

$\begin{aligned}
& \text {L.H.S. }=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x} \\
& =\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x} \\
& =\frac{2 \cos \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\cos 3 x}{2 \sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\sin 3 x} \\
& =\frac{2 \cos 3 x \cos x+\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x} \\
& =\frac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)}=\cot 3 x=\text { R.H.S. }
\end{aligned}$
Ex 3.3 Question 22.

$\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$

Answer.

We know that $\cot 3 x=\cot (2 x+x)$

$
\begin{aligned}
& \Rightarrow \cot 3 x=\frac{\cot 2 x \cot x-1}{\cot 2 x+\cot x} \\
& \Rightarrow \cot 3 x(\cot 2 x+\cot x)=\cot 2 x \cot x-1 \\
& \Rightarrow \cot 3 x \cot 2 x+\cot 3 x \cot x=\cot 2 x \cot x-1 \\
& \Rightarrow \cot 3 x \cot 2 x+\cot 3 x \cot x-\cot 2 x \cot x+1=0 \\
& \Rightarrow \cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1
\end{aligned}
$
Ex 3.3 Question 23.

$\tan 4 x=\frac{4 \tan x\left(1-\tan ^2 x\right)}{1-6 \tan ^2 x+\tan ^4 x}$

Answer.

L.H.S. $=\tan 4 x=\frac{2 \tan 2 x}{1-\tan ^2 2 x}$
$
=\frac{2 \cdot \frac{2 \tan x}{1-\tan ^2 x}}{1-\left(\frac{2 \tan x}{1-\tan ^2 x}\right)^2}
$

$
\begin{aligned}
& =\frac{\frac{4 \tan x}{1-\tan ^2 x}}{\frac{\left(1-\tan ^2 x\right)^2-4 \tan ^2 x}{\left(1-\tan ^2 x\right)^2}} \\
& =\frac{4 \tan x}{1-\tan ^2 x} \times \frac{\left(1-\tan ^2 x\right)^2}{1+\tan ^4 x-2 \tan ^2 x-4 \tan ^2 x} \\
& =\frac{4 \tan x\left(1-\tan ^2 x\right)}{1-6 \tan ^2 x+\tan ^4 x}=\text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 24.

$\cos 4 x=1-8 \sin ^2 x \cos ^2 x$

Answer:

$
\begin{aligned}
& \text {L.H.S. }=\cos 4 x=1-2 \sin ^2 2 x \\
& =1-2(2 \sin x \cos x)^2 \\
& =1-2\left(4 \sin ^2 x \cos ^2 x\right) \\
& =1-8 \sin ^2 x \cos ^2 x=\text { R.H.S. }
\end{aligned}
$
Ex 3.3 Question 25.

$\cos 6 x=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1$

Answer:

$
\begin{aligned}
& \text {L.H.S. }=\cos 6 x=2 \cos ^2 3 x-1 \\
& =2\left[4 \cos ^3 x-3 \cos x\right]^2-1 \\
& =2\left[16 \cos ^6 x+9 \cos ^2 x-24 \cos ^4 x\right]-1 \\
& =32 \cos ^6 x+18 \cos ^2 x-48 \cos ^4 x-1 \\
& =32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1=\text { R.H.S. }
\end{aligned}
$