Answer :

L.H.S. $=\sin ^{2} \dfrac{\pi}{6}+\cos ^{2} \dfrac{\pi}{3}-\tan ^{2} \dfrac{\pi}{4}$

$=(\dfrac{1}{2})^{2}+(\dfrac{1}{2})^{2}-(1)^{2}$

$=\dfrac{1}{4}+\dfrac{1}{4}-1=-\dfrac{1}{2}$

$=$ R.H.S.

Answer :

L.H.S. $=2 \sin ^{2} \dfrac{\pi}{6}+cosec 2 \dfrac{7 \pi}{6} \cos ^{2} \dfrac{\pi}{3}$

$ \begin{aligned} & =2(\dfrac{1}{2})^{2}+cosec^{2}(\pi+\dfrac{\pi}{6})(\dfrac{1}{2})^{2} \\ & =2 \times \dfrac{1}{4}+(-cosec \dfrac{\pi}{6})^{2}(\dfrac{1}{4}) \\ & =\dfrac{1}{2}+(-2)^{2}(\dfrac{1}{4}) \\ & =\dfrac{1}{2}+\dfrac{4}{4}=\dfrac{1}{2}+1=\dfrac{3}{2} \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

L.H.S. $=\cot ^{2} \dfrac{\pi}{6}+cosec \dfrac{5 \pi}{6}+3 \tan ^{2} \dfrac{\pi}{6}$

$=(\sqrt{3})^{2}+cosec(\pi-\dfrac{\pi}{6})+3(\dfrac{1}{\sqrt{3}})^{2}$

$=3+cosec \dfrac{\pi}{6}+3 \times \dfrac{1}{3}$

$=3+2+1=6$

$=$ R.H.S

Answer :

L.H.S $=2 \sin ^{2} \dfrac{3 \pi}{4}+2 \cos ^{2} \dfrac{\pi}{4}+2 \sec ^{2} \dfrac{\pi}{3}$ $=2\{\sin (\pi-\dfrac{\pi}{4})\}^{2}+2(\dfrac{1}{\sqrt{2}})^{2}+2(2)^{2}$

$=2\{\sin \dfrac{\pi}{4}\}^{2}+2 \times \dfrac{1}{2}+8$

$=2(\dfrac{1}{\sqrt{2}})^{2}+1+8$

$=1+1+8$

$=10$

$=$ R.H.S

Answer :

(i) $\sin 75^{\circ}=\sin (45^{\circ}+30^{\circ})$

$=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}$

$[\sin (x+y)=\sin x \cos y+\cos x \sin y]$

$=(\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{3}}{2})+(\dfrac{1}{\sqrt{2}})(\dfrac{1}{2})$

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

(ii) $\tan 15^{\circ}=\tan (45^{\circ} \hat{a} \in^{\prime \prime} 30^{\circ})$

$ \begin{aligned} & =\dfrac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} \quad[\tan (x-y)=\dfrac{\tan x-\tan y}{1+\tan x \tan y}] \\ & =\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+1(\dfrac{1}{\sqrt{3}})}=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}} \\ & =\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\dfrac{3+1-2 \sqrt{3}}{(\sqrt{3})^{2}-(1)^{2}} \\ & =\dfrac{4-2 \sqrt{3}}{3-1}=2-\sqrt{3} \end{aligned} $

Prove the following:

Answer :

$ \begin{aligned} & \cos (\dfrac{\pi}{4}-x) \cos (\dfrac{\pi}{4}-y)-\sin (\dfrac{\pi}{4}-x) \sin (\dfrac{\pi}{4}-y) \\ &= \dfrac{1}{2}[2 \cos (\dfrac{\pi}{4}-x) \cos (\dfrac{\pi}{4}-y)]+\dfrac{1}{2}[-2 \sin (\dfrac{\pi}{4}-x) \sin (\dfrac{\pi}{4}-y)] \\ &= \dfrac{1}{2}[\cos \{(\dfrac{\pi}{4}-x)+(\dfrac{\pi}{4}-y)\}+\cos \{(\dfrac{\pi}{4}-x)-(\dfrac{\pi}{4}-y)\}] \\ &+\dfrac{1}{2}[\cos \{(\dfrac{\pi}{4}-x)+(\dfrac{\pi}{4}-y)\}-\cos \{(\dfrac{\pi}{4}-x)-(\dfrac{\pi}{4}-y)\}] \\ & {[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] } \\ &-2 \sin A \sin B=\cos (A+B)-\cos (A-B)] \\ &= 2 \times \dfrac{1}{2}[\cos \{(\dfrac{\pi}{4}-x)+(\dfrac{\pi}{4}-y)\}] \\ &= \cos [\dfrac{\pi}{2}-(x+y)] \\ &= \sin (x+y) \\ &= \text{ R.H.S } \end{aligned} $

Answer :

It is known that

$ \tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A \tan B} \text{ and } \tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A \tan B} $

$ \dfrac{\tan (\dfrac{\pi}{4}+x)}{\tan (\dfrac{\pi}{4}-x)}=\dfrac{(\dfrac{\tan \dfrac{\pi}{4}+\tan x}{1-\tan \dfrac{\pi}{4} \tan x})}{(\dfrac{\tan \dfrac{\pi}{4}-\tan x}{1+\tan \dfrac{\pi}{4} \tan x})}=\dfrac{(\dfrac{1+\tan x}{1-\tan x})}{(\dfrac{1-\tan x}{1+\tan x})}=(\dfrac{1+\tan x}{1-\tan x})^{2}=\text{ R.H.S. } $

Answer :

$ \begin{aligned} \text{ L.H.S. } & =\dfrac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos (\dfrac{\pi}{2}+x)} \\ & =\dfrac{[-\cos x][\cos x]}{(\sin x)(-\sin x)} \\ & =\dfrac{-\cos ^{2} x}{-\sin ^{2} x} \\ & =\cot ^{2} x \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

$ \begin{aligned} & \text{ L.H.S. }=\cos (\dfrac{3 \pi}{2}+x) \cos (2 \pi+x)[\cot (\dfrac{3 \pi}{2}-x)+\cot (2 \pi+x)] \\ & =\sin x \cos x[\tan x+\cot x] \\ & =\sin x \cos x(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}) \\ & =(\sin x \cos x)[\dfrac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}] \\ & =1=\text{ R.H.S. } \end{aligned} $

Answer :

L.H.S. $=\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x$

$=\dfrac{1}{2}[2 \sin (n+1) x \sin (n+2) x+2 \cos (n+1) x \cos (n+2) x]$

$=\dfrac{1}{2} \begin{cases} \cos \{(n+1) x-(n+2) x\}-\cos \{(n+1) x+(n+2) x\} \\ +\cos \{(n+1) x+(n+2) x\}+\cos \{(n+1) x-(n+2) x\} \end{cases} $

$ \begin{cases} \because-2 \sin A \sin B=\cos (A+B)-\cos (A-B) \\ 2 \cos A \cos B=\cos (A+B)+\cos (A-B) \end{cases} $

$=\dfrac{1}{2} \times 2 \cos \{(n+1) x-(n+2) x\}$

$=\cos (-x)=\cos x=$ R.H.S.

Answer :

It is known that

$ \cos A-\cos B=-2 \sin (\dfrac{A+B}{2}) \cdot \sin (\dfrac{A-B}{2}) $

$\therefore$ L.H.S. $=\cos (\dfrac{3 \pi}{4}+x)-\cos (\dfrac{3 \pi}{4}-x)$

$=-2 \sin \{\dfrac{(\dfrac{3 \pi}{4}+x)+(\dfrac{3 \pi}{4}-x)}{2}\} \cdot \sin \{\dfrac{(\dfrac{3 \pi}{4}+x)-(\dfrac{3 \pi}{4}-x)}{2}\}$

$=-2 \sin (\dfrac{3 \pi}{4}) \sin x$

$=-2 \sin (\pi-\dfrac{\pi}{4}) \sin x$

$=-2 \sin \dfrac{\pi}{4} \sin x$

$=-2 \times \dfrac{1}{\sqrt{2}} \times \sin x$

$=-\sqrt{2} \sin x$

$=$ R.H.S.

Answer :

It is known

that $\sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2}), \sin A-\sin B=2 \cos (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2})$

$\therefore$ L.H.S. $=\sin ^{2} 6 x$ -$\sin ^{2} 4 x$

$=(\sin 6 x+\sin 4 x)(\sin 6 x$ -$\sin$

$4 x)$

$=[2 \sin (\dfrac{6 x+4 x}{2}) \cos (\dfrac{6 x-4 x}{2})][2 \cos (\dfrac{6 x+4 x}{2}) \cdot \sin (\dfrac{6 x-4 x}{2})]$

$ \begin{aligned} & =(2 \sin 5 x \cos x)(2 \cos 5 x \sin x) \\ & =(2 \sin 5 x \cos 5 x)(2 \sin x \cos x) \\ & =\sin 10 x \sin 2 x \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

L.H.S. $=\sin 2 x+2 \sin 4 x+\sin 6 x$

$=[\sin 2 x+\sin 6 x]+2 \sin 4 x$

$=[2 \sin (\dfrac{2 x+6 x}{2}) \cos (\dfrac{2 x-6 x}{2})]+2 \sin 4 x$

$[\because \sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})]$

$=2 \sin 4 x \cos ( 2 x)+2 \sin 4 x$

$ \begin{aligned} & =2 \sin 4 x \cos 2 x+2 \sin 4 x \\ & =2 \sin 4 x(\cos 2 x+1) \\ & =2 \sin 4 x(2 \cos ^{2} x - 1+1) \\ & =2 \sin 4 x(2 \cos ^{2} x) \\ & =4 \cos ^{2} x \sin 4 x \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

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

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

$[\because \sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})]$

$=(\dfrac{\cos 4 x}{\sin 4 x})[2 \sin 4 x \cos x]$

$=2 \cos 4 x \cos x$

R.H.S. $=\cot x(\sin 5 x$ - $\sin 3 x)$

$=\dfrac{\cos x}{\sin x}[2 \cos (\dfrac{5 x+3 x}{2}) \sin (\dfrac{5 x-3 x}{2})]$

$[\because \sin A-\sin B=2 \cos (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2})]$

$=\dfrac{\cos x}{\sin x}[2 \cos 4 x \sin x]$

$=2 \cos 4 x \cdot \cos x$

L.H.S. = R.H.S.

Answer :

It is known that

$ \begin{aligned} & \cos A-\cos B=-2 \sin (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2}), \sin A-\sin B=2 \cos (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2}) \\ & \therefore \text{ L.H.S }=\dfrac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x} \\ & =\dfrac{-2 \sin (\dfrac{9 x+5 x}{2}) \cdot \sin (\dfrac{9 x-5 x}{2})}{2 \cos (\dfrac{17 x+3 x}{2}) \cdot \sin (\dfrac{17 x-3 x}{2})} \\ & =\dfrac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin 7 x} \\ & =-\dfrac{\sin 2 x}{\cos 10 x} \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

It is known that

$ \sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2}), \cos A+\cos B=2 \cos (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2}) $

$\therefore$ L.H.S. $=\dfrac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$

$=\dfrac{2 \sin (\dfrac{5 x+3 x}{2}) \cdot \cos (\dfrac{5 x-3 x}{2})}{2 \cos (\dfrac{5 x+3 x}{2}) \cdot \cos (\dfrac{5 x-3 x}{2})}$

$=\dfrac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x}$

$=\dfrac{\sin 4 x}{\cos 4 x}$

$=\tan 4 x=$ R.H.S.

Answer :

It is known that

$\sin A-\sin B=2 \cos (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2}), \cos A+\cos B=2 \cos (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})$

$\therefore$ L.H.S. $=\dfrac{\sin x-\sin y}{\cos x+\cos y}$

$=\dfrac{2 \cos (\dfrac{x+y}{2}) \cdot \sin (\dfrac{x-y}{2})}{2 \cos (\dfrac{x+y}{2}) \cdot \cos (\dfrac{x-y}{2})}$

$=\dfrac{\sin (\dfrac{x-y}{2})}{\cos (\dfrac{x-y}{2})}$

$=\tan (\dfrac{x-y}{2})=$ R.H.S.

Answer :

It is known that

$\sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2}), \cos A+\cos B=2 \cos (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})$

$\therefore$ L.H.S. $=\dfrac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$

$ \begin{aligned} & =\dfrac{2 \sin (\dfrac{x+3 x}{2}) \cos (\dfrac{x-3 x}{2})}{2 \cos (\dfrac{x+3 x}{2}) \cos (\dfrac{x-3 x}{2})} \\ & =\dfrac{\sin 2 x}{\cos 2 x} \\ & =\tan 2 x \\ & =\text{ R.H.S } \end{aligned} $

Answer :

It is known that

$ \sin A-\sin B=2 \cos (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2}), \cos ^{2} A-\sin ^{2} A=\cos 2 A $

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

$=\dfrac{2 \cos (\dfrac{x+3 x}{2}) \sin (\dfrac{x-3 x}{2})}{-\cos 2 x}$

$=\dfrac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$

$=-2 \times(-\sin x)$

$=2 \sin x=$ R.H.S.

Answer :

L.H.S. $=\dfrac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$

$ \begin{aligned} & =\dfrac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x} \\ & =\dfrac{2 \cos (\dfrac{4 x+2 x}{2}) \cos (\dfrac{4 x-2 x}{2})+\cos 3 x}{2 \sin (\dfrac{4 x+2 x}{2}) \cos (\dfrac{4 x-2 x}{2})+\sin 3 x} \\ & {[\because \cos A+\cos B=2 \cos (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2}), \sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})]} \\ & =\dfrac{2 \cos 3 x \cos x+\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x} \\ & =\dfrac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)} \\ & =\cot 3 x=\text{ R.H.S. } \end{aligned} $

Answer :

L.H.S. $=\cot x \cot 2 x$ -$\cot 2 x \cot 3 x$ -$\cot 3 x \cot x$

$=\cot x \cot 2 x$ -$\cot 3 x(\cot 2 x+\cot x)$

$=\cot x \cot 2 x - \cot (2 x+x)(\cot 2 x+\cot x)$

$=\cot x \cot 2 x-[\dfrac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}](\cot 2 x+\cot x)$

$[\because \cot (A+B)=\dfrac{\cot A \cot B-1}{\cot A+\cot B}]$

$=\cot x \cot 2 x$ -( $\cot 2 x \cot x$ -1$)$

$=1=$ R.H.S.

Answer :

$ \begin{aligned} & \text{ It is known that } \tan 2 A=\dfrac{2 \tan A}{1-\tan ^{2} A} \\ & \therefore \text{ L.H.S. }=\tan 4 x=\tan 2(2 x) \\ & =\dfrac{2 \tan 2 x}{1-\tan ^{2}(2 x)} \\ & =\dfrac{2(\dfrac{2 \tan x}{1-\tan ^{2} x})}{1-(\dfrac{2 \tan x}{1-\tan ^{2} x})^{2}} \\ & =\dfrac{(\dfrac{4 \tan x}{1-\tan ^{2} x})}{[1-\dfrac{4 \tan ^{2} x}{(1-\tan ^{2} x)^{2}}]} \\ & =\dfrac{(\dfrac{4 \tan x}{1-\tan ^{2} x})}{[\dfrac{(1-\tan ^{2} x)^{2}-4 \tan ^{2} x}{(1-\tan ^{2} x)^{2}}]} \\ & =\dfrac{4 \tan x(1-\tan ^{2} x)}{(1-\tan ^{2} x)^{2}-4 \tan ^{2} x} \\ & =\dfrac{4 \tan x(1-\tan ^{2} x)}{1+\tan ^{4} x-2 \tan ^{2} x-4 \tan ^{2} x} \\ & =\dfrac{4 \tan x(1-\tan ^{2} x)}{1-6 \tan ^{2} x+\tan ^{4} x}=\text{ R.H.S. } \end{aligned} $

Answer :

L.H.S. $=\cos 6 x$

$=\cos 3(2 x)$

$=4 \cos ^{3} 2 x-3 \cos 2 x[\cos 3 A=4 \cos ^{3} A-3 \cos A]$

$=4[(2 \cos ^{2} x-1)^{3}-3(2 \cos ^{2} x-1)[\cos 2 x=2 \cos ^{2} x-1].$

$=4[(2 \cos ^{2} x)^{3}-(1)^{3}-3(2 \cos ^{2} x)^{2}+3(2 \cos ^{2} x)]-6 \cos ^{2} x+3$

$=4[8 \cos ^{6} x-1-12 \cos ^{4} x+6 \cos ^{2} x]-6 \cos ^{2} x+3$

$=32 \cos ^{6} x-4-48 \cos ^{4} x+24 \cos ^{2} x-6 \cos ^{2} x+3$

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

$=$ R.H.S.