Prove that
1.$\sin^{2}\frac{\pi }{6}+\cos^{2}\frac{\pi }{3}-\tan^{2}\frac{\pi }{4}=-\frac{1}{2}$
L.H.S.,
= $\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{2}-1$
= $\frac{1}{4}+\frac{1}{4}-1$
=$\frac{1}{2}-1$
=$-\frac{1}{2}$
L.H.S.=R.H.S.
2. $2\sin^{2}\frac{\pi }{6}+\csc ^{2}\frac{7\pi }{6}\cos^{2}\frac{\pi }{3}$
L.H.S.,
= $2\left ( \frac{1}{2} \right )^{2}+\csc ^{2}\left (\pi +\frac{\pi }{6} \right )\left ( \frac{1}{2} \right )^{2}$
= $2\left ( \frac{1}{4} \right )+\left ( -\csc\frac{\pi }{6} \right )^{2}\frac{1}{4}$
= $\frac{1}{2}+\left ( -2 \right )^{2}\frac{1}{4}$
= $\frac{1}{2}+1$
= $\frac{3}{2}$
L.H.S.=R.H.S.
3. $\cot^{2}\frac{\pi }{6}+\csc \frac{5\pi }{6}+3\tan^{2}\frac{\pi }{6}=6$
L.H.S.,
= $\left ( \sqrt{3} \right )^{2}+\csc \left ( \pi -\frac{\pi }{6} \right )+3\left ( \frac{1}{\sqrt{3}} \right )^{2}$
= $3+\csc \frac{\pi }{6}+3*\frac{1}{3}$
= $3+2+1=6$
L.H.S.=R.H.S.
4. $2\sin^{2}\frac{3\pi}{4}+2\cos^{2}\frac{\pi}{4}+2\sec^{2}\frac{\pi}{3}=10$
L.H.S.,
= $2\sin^{2}\left ( \pi-\frac{\pi}{4} \right )+2\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( 2 \right )^{2}$
= $2\sin^{2}\left (\frac{\pi}{4} \right )+2\left ( \frac{1}{2} \right )+2\left ( 4 \right )$
= $2\left ( \frac{1}{\sqrt{2}} \right )^{2}+1+8$
= $2\left ( \frac{1}{2} \right )+9=10$
L.H.S.=R.H.S.
5. Find the value of
(i) sin 75°
= sin(45°+30°)
= sin 45° cos 30° + cos 45° sin 30° [sin(x+y)=sin x cos y + cos x sin y]
= $\left ( \frac{1}{\sqrt{2}} \right )\left ( \frac{\sqrt{3}}{2} \right )+\left ( \frac{1}{\sqrt{2}} \right )\left ( \frac{1}{2} \right )$
= $\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$
(ii) tan 15°
= tan(45°-30°)
= $\large \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}$ [$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$ ]
= $\large \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )}$
= $\large \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$
= $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
= $\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}$
= $\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3} \right )^{2}-1^{2}}$
= $\frac{\left ( \sqrt{3} \right )^{2}+1-2\sqrt{3}}{3-1}$
= $\frac{3+1-2\sqrt{3}}{2}$
= $\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
Prove the following:
6. $\cos \left ( \frac{\pi}{4}-x \right )$ $\cos \left ( \frac{\pi}{4}-y \right )$ $-\sin \left ( \frac{\pi}{4}-x \right )$ $\sin \left ( \frac{\pi}{4}-y \right )$ $=\sin (x+y)$
L.H.S.,
= $\cos \left [ \left ( \frac{\pi}{4}-x \right )+ \left ( \frac{\pi}{4}-y \right ) \right ]$ [cos(x+y) = cos x cos y-sin x sin y]
= $\cos\left ( 2\frac{\pi}{4}-x-y \right )$
= $\cos\left ( \frac{\pi}{2}-(x+y) \right )$
= $\sin(x+y)$
L.H.S.=R.H.S.
7. $\large \frac{\tan\left ( \frac{\pi}{4}+x \right )}{\tan\left ( \frac{\pi}{4}-x \right )}=\left ( \frac{1+\tan x}{1-\tan x} \right )^{2}$
We know that,
$\tan (x+y)= \frac{\tan x+\tan y}{1-\tan x\tan y}$
$\tan (x-y)= \frac{\tan x-\tan y}{1+\tan x\tan y}$
L.H.S.,
= $\frac{\tan\left ( \frac{\pi}{4}+x \right )}{\tan\left ( \frac{\pi}{4}-x \right )}$
= $\LARGE \frac{\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}}{\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}}$
= $\LARGE \frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}$
= $\frac{1+\tan x}{1-\tan x}*\frac{1+\tan x}{1-\tan x}$
= $\left (\frac{1+\tan x}{1-\tan x} \right )^{2}$
L.H.S.=R.H.S.
8. $\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)}=\cot^{2}x$
L.H.S.,
= $\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$
= $\frac{\cos^{2}x}{\sin^{2}x}=\cot^{2}x$
L.H.S.=R.H.S.
9. $\cos \left ( \frac{3\pi}{2}+x \right )$ $\cos \left ( 2\pi+x \right )$ [$\cot \left ( \frac{3\pi}{2}+x \right )$+ $\cot \left ( 2\pi+x \right )$]=1
L.H.S.,
= sin x cos x [tan x + cot x]
= $\sin x\cos x\left [ \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \right ]$
= $\sin x\cos x\left [ \frac{\sin^{2} x+\cos^{2}x}{\cos x\sin x}\right ]$
= $\sin^{2} x+\cos^{2}x$ =1
L.H.S.=R.H.S.
10. sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x = cos x
L.H.S.,
= cos[(n+1)x-(n+2)x] [cos(x-y)=cos x cos y+sin x sin y]
= cos(n+1-n-2)x
= cos (-x) = cos x
L.H.S.=R.H.S.
11. $\cos \left ( \frac{3\pi}{4}+x \right )$-$\cos \left ( \frac{3\pi}{4}-x \right )$$=-\sqrt{2}\sin x$
We know that,
cos(x+y)=cos x cos y-sin x sin y
cos(x-y)=cos x cos y+sin x sin y
L.H.S.,
=$\cos \frac{3\pi}{4}\cos x$ $-\sin \frac{3\pi}{4}\sin x$ $-\cos \frac{3\pi}{4}\cos x$$-\sin \frac{3\pi}{4}\sin x$
= $-2\sin \frac{3\pi}{4}\sin x$
= $-2\sin \left ( \pi-\frac{\pi}{4} \right )\sin x$
= $-2\sin \left ( \frac{\pi}{4} \right )\sin x$
= $-2\left ( \frac{1}{\sqrt{2}} \right )\sin x$ = $-\sqrt{2}\sin x$
L.H.S.=R.H.S.
12. $\sin^{2}6x-\sin^{2}4x=\sin 2x \sin 10x$
We know that,
$\sin x+\sin y=2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\sin x-\sin y=2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
L.H.S.,
= (sin 6x-sin 4x)(sin 6x+sin 4x)
= $\left [ 2\cos\left ( \frac{6x+4x}{2} \right )\sin\left ( \frac{6x-4x}{2} \right ) \right ]$ $\left [ 2\sin\left ( \frac{6x+4x}{2} \right )\cos\left ( \frac{6x-4x}{2} \right ) \right ]$
= (2 cos 5x sin x)(2 sin 5x cos x)
= (2 cos 5x sin 5x) (2 cos x sin x)
= cos 10x sin 2x [sin 2x=2 sin x cos x]
L.H.S.=R.H.S.
13. $\cos^{2}2x-\cos^{2}6x=\sin 4x\sin 8x$
We know that,
$\cos x+\cos y=2\cos\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\cos x-\cos y=-2\sin\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
L.H.S.,
= (cos 2x-cos 6x)(cos 2x+cos 6x)
=$\left [ -2\sin\left ( \frac{2x+6x}{2} \right )\sin\left ( \frac{2x-6x}{2} \right ) \right ]$ $\left [ 2\cos\left ( \frac{2x+6x}{2} \right )\cos\left ( \frac{2x-6x}{2} \right ) \right ]$
= [-2 sin 4x sin(-2x)][2 cos 4x cos(-2x)]
= (2 sin 4x sin 2x) (2 cos 4x cos 2x)
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x [sin 2x=2 sin x cos x]
L.H.S.=R.H.S.
14. $\sin 2x+2\sin 4x$ $+\sin 6x$ $=4\cos^{2}x\sin 4x$
L.H.S.,
= [sin 2x + sin 6x]+2sin 4x
= $\left [ 2\sin\left ( \frac{2x+6x}{2} \right )\cos\left ( \frac{2x-6x}{2} \right ) \right ]+2\sin 4x$ [$\sin x+\sin y=2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$]
= [2 sin 4x cos(-2x)]+2sin 4x
= 2 sin 4x cos 2x + 2sin 4x
= 2 sin 4x [cos 2x + 1]
= 2 sin 4x [2cos$\displaystyle \small ^{2}$ x-1+1] [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
= 2 sin 4x (2cos$\displaystyle \small ^{2}$ x) = 4 cos$\displaystyle \small ^{2}$ x sin 4x
L.H.S.=R.H.S.
15. $\cot 4x\left ( \sin 5x+\sin 3x \right )$ $=\cot x\left ( \sin 5x-\sin 3x \right )$
L.H.S.,
=$\frac{\cos 4x}{\sin 4x}\left [ 2\sin\left ( \frac{5x+3x}{2} \right )\cos\left ( \frac{5x-3x}{2} \right ) \right ]$ [$\sin x+\sin y=2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$]
= $\frac{\cos 4x}{\sin 4x}\left [ 2\sin 4x\cos x \right ]$
= $2\cos 4x \cos x$
R.H.S.,
= $\frac{\cos x}{\sin x}\left [ 2\cos\left ( \frac{5x+3x}{2} \right )\sin\left ( \frac{5x-3x}{2} \right ) \right ]$ [$\sin x-\sin y=2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$]
= $\frac{\cos x}{\sin x}\left [ 2\cos 4x\sin x \right ]$
= $2\cos 4x \cos x$
L.H.S.=R.H.S.
16. $\large \frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\frac{\sin 2x}{\cos 10x}$
L.H.S.,
= $\large \frac{-2\sin\left ( \frac{9x+5x}{2} \right )\sin\left ( \frac{9x-5x}{2} \right )}{2\cos\left ( \frac{17x+3x}{2} \right )\sin\left ( \frac{17x-3x}{2} \right )}$
= $\frac{-2\sin 7x \sin 2x}{2 \cos 10x\sin 7x}$
= $-\frac{ \sin 2x}{ \cos 10x}$
L.H.S.=R.H.S.
17. $\large \frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$
L.H.S.,
= $\large \frac{2\sin\left ( \frac{5x+3x}{2} \right )\cos\left ( \frac{5x-3x}{2} \right )}{2\cos\left ( \frac{5x+3x}{2} \right )\cos\left ( \frac{5x-3x}{2} \right )}$
= $\frac{2\sin 4x \cos x}{2 \cos 4x\cos x}$
= $\tan 4x$
L.H.S.=R.H.S.
18. $\large \frac{\sin x-\sin y}{\cos x+\cos y}=\tan\left ( \frac{x-y}{2} \right )$
L.H.S.,
= $\large \frac{2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )}{2\cos\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )}$
= $\tan\left ( \frac{x-y}{2} \right )$
L.H.S.=R.H.S.
19. $\large \frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 4x$
L.H.S.,
= $\large \frac{2\sin\left ( \frac{x+3x}{2} \right )\cos\left ( \frac{x-3x}{2} \right )}{2\cos\left ( \frac{x+3x}{2} \right )\cos\left ( \frac{x-3x}{2} \right )}$
= $\frac{\sin 2x }{ \cos 2x}$
= $\tan 2x$
L.H.S.=R.H.S.
20. $\large \frac{\sin x-\sin 3x}{\sin^{2}x-\cos^{2}x}=2\sin x$
L.H.S.,
= $\large \frac{2\cos\left ( \frac{x+3x}{2} \right )\sin\left ( \frac{x-3x}{2} \right )}{-\cos 2x}$
= $\large -\frac{2 \cos 2x \sin (-x)}{\cos 2x}$
= $\large 2\sin x$
L.H.S.=R.H.S.
21. $\large \frac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$
L.H.S.,
= $\frac{\left (\cos 4x+\cos 2x \right )+\cos 3x}{\left (\sin 4x+\sin 2x \right )+\sin 3x}$
= $\large \frac{2\cos\left ( \frac{4x+2x}{2} \right )\cos\left ( \frac{4x-2x}{2} \right )+\cos 3x}{2\sin\left ( \frac{4x+2x}{2} \right )\cos\left ( \frac{4x-2x}{2} \right )+\sin 3x}$
= $\frac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x}$
= $\frac{\cos 3x(2\cos x+1)}{\sin 3x(2\cos x+1)}$
= $\cot 3x$
L.H.S.=R.H.S.
22. $\cot x\cot 2x$ $-\cot 2x\cot 3x$ $-\cot 3x\cot x=1$
L.H.S.,
= $\cot x\cot 2x-\cot 3x(\cot 2x+\cot x)$
= $\cot x\cot 2x$ $-\cot (2x+x)(\cot 2x+\cot x)$
= $\cot x\cot 2x$ $-\left [ \frac{\cot 2x\cot x-1}{\cot 2x+\cot x} \right ]$ $(\cot 2x+\cot x)$ [$\cot(x+y)=\frac{\cot x\cot y-1}{\cot x+\cot y}$ ]
= $\cot x\cot 2x-\cot 2x\cot x+1$ = 1
L.H.S.=R.H.S.
23. $\tan 4x=\frac{4\tan x(1-\tan^{2}x)}{1-6\tan^{2}x+\tan^{4}x}$
L.H.S.,
= $\tan 2(2x)$
= $\frac{2\tan 2x}{1-\tan^{2}2x}$ [$\tan 2x=\frac{2\tan x}{1-\tan^{2}x}$ ]
= $\large \frac{2\left (\frac{2\tan x}{1-\tan^{2}x} \right )}{1-\left (\frac{2\tan x}{1-\tan^{2}x} \right )^{2}}$
= $\large \frac{2\left (\frac{2\tan x}{1-\tan^{2}x} \right )}{1-\frac{4\tan^{2}x}{(1-\tan^{2}x)^{2}}}$
= $\large \frac{2\left (\frac{2\tan x}{1-\tan^{2}x} \right )}{\frac{(1-\tan^{2}x)^{2}-4\tan^{2}x}{(1-\tan^{2}x)^{2}}}$
= $\frac{4\tan x}{1-\tan^{2}x}*\frac{(1-\tan^{2}x)^{2}}{(1-\tan^{2}x)^{2}-4\tan^{2}x}$
= $\frac{4\tan x(1-\tan^{2}x)}{1+\tan^{4}x-2\tan^{2}x-4\tan^{2}x}$
= $\frac{4\tan x(1-\tan^{2}x)}{1+\tan^{4}x-6\tan^{2}x}$
L.H.S.=R.H.S.
24. $\cos 4x=1-8\sin^{2}x\cos^{2}x$
L.H.S.,
= $\cos 2(2x)$
= $1-2\sin^{2}2x$ [cos 2x=1-2sin$\displaystyle \small ^{2}$ x]
= $1-2(2\sin x\cos x)^{2}$
= $1-8\sin^{2} x\cos^{2} x$
L.H.S.=R.H.S.
25. $\cos 6x=32\cos^{6}x$ $-48\cos^{4}x+18\cos^{2}x-1$
L.H.S.,
= $\cos 3(2x)$
= $4\cos^{3}2x-3\cos 2x$ [cos 3x=4cos$\displaystyle \small ^{3}$ x-3cosx]
= $4[2\cos^{2}x-1]^{3}-3[2\cos^{2}x-1]$
= $4[(2\cos^{2}x)^{3}-1^{3}-3(2\cos^{2}x)^{2}$ $+3(2\cos^{2}x)]-6\cos^{2}x+3$
= $32\cos^{6}x-48\cos^{4}x+24\cos^{2}x$ $-4-6\cos^{2}x+3$
= $32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1$
L.H.S.=R.H.S.
1.$\sin^{2}\frac{\pi }{6}+\cos^{2}\frac{\pi }{3}-\tan^{2}\frac{\pi }{4}=-\frac{1}{2}$
L.H.S.,
= $\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{2}-1$
= $\frac{1}{4}+\frac{1}{4}-1$
=$\frac{1}{2}-1$
=$-\frac{1}{2}$
L.H.S.=R.H.S.
2. $2\sin^{2}\frac{\pi }{6}+\csc ^{2}\frac{7\pi }{6}\cos^{2}\frac{\pi }{3}$
L.H.S.,
= $2\left ( \frac{1}{2} \right )^{2}+\csc ^{2}\left (\pi +\frac{\pi }{6} \right )\left ( \frac{1}{2} \right )^{2}$
= $2\left ( \frac{1}{4} \right )+\left ( -\csc\frac{\pi }{6} \right )^{2}\frac{1}{4}$
= $\frac{1}{2}+\left ( -2 \right )^{2}\frac{1}{4}$
= $\frac{1}{2}+1$
= $\frac{3}{2}$
L.H.S.=R.H.S.
3. $\cot^{2}\frac{\pi }{6}+\csc \frac{5\pi }{6}+3\tan^{2}\frac{\pi }{6}=6$
L.H.S.,
= $\left ( \sqrt{3} \right )^{2}+\csc \left ( \pi -\frac{\pi }{6} \right )+3\left ( \frac{1}{\sqrt{3}} \right )^{2}$
= $3+\csc \frac{\pi }{6}+3*\frac{1}{3}$
= $3+2+1=6$
L.H.S.=R.H.S.
4. $2\sin^{2}\frac{3\pi}{4}+2\cos^{2}\frac{\pi}{4}+2\sec^{2}\frac{\pi}{3}=10$
L.H.S.,
= $2\sin^{2}\left ( \pi-\frac{\pi}{4} \right )+2\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( 2 \right )^{2}$
= $2\sin^{2}\left (\frac{\pi}{4} \right )+2\left ( \frac{1}{2} \right )+2\left ( 4 \right )$
= $2\left ( \frac{1}{\sqrt{2}} \right )^{2}+1+8$
= $2\left ( \frac{1}{2} \right )+9=10$
L.H.S.=R.H.S.
5. Find the value of
(i) sin 75°
= sin(45°+30°)
= sin 45° cos 30° + cos 45° sin 30° [sin(x+y)=sin x cos y + cos x sin y]
= $\left ( \frac{1}{\sqrt{2}} \right )\left ( \frac{\sqrt{3}}{2} \right )+\left ( \frac{1}{\sqrt{2}} \right )\left ( \frac{1}{2} \right )$
= $\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$
(ii) tan 15°
= tan(45°-30°)
= $\large \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}$ [$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$ ]
= $\large \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )}$
= $\large \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$
= $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
= $\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}$
= $\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3} \right )^{2}-1^{2}}$
= $\frac{\left ( \sqrt{3} \right )^{2}+1-2\sqrt{3}}{3-1}$
= $\frac{3+1-2\sqrt{3}}{2}$
= $\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
Prove the following:
6. $\cos \left ( \frac{\pi}{4}-x \right )$ $\cos \left ( \frac{\pi}{4}-y \right )$ $-\sin \left ( \frac{\pi}{4}-x \right )$ $\sin \left ( \frac{\pi}{4}-y \right )$ $=\sin (x+y)$
L.H.S.,
= $\cos \left [ \left ( \frac{\pi}{4}-x \right )+ \left ( \frac{\pi}{4}-y \right ) \right ]$ [cos(x+y) = cos x cos y-sin x sin y]
= $\cos\left ( 2\frac{\pi}{4}-x-y \right )$
= $\cos\left ( \frac{\pi}{2}-(x+y) \right )$
= $\sin(x+y)$
L.H.S.=R.H.S.
7. $\large \frac{\tan\left ( \frac{\pi}{4}+x \right )}{\tan\left ( \frac{\pi}{4}-x \right )}=\left ( \frac{1+\tan x}{1-\tan x} \right )^{2}$
We know that,
$\tan (x+y)= \frac{\tan x+\tan y}{1-\tan x\tan y}$
$\tan (x-y)= \frac{\tan x-\tan y}{1+\tan x\tan y}$
L.H.S.,
= $\frac{\tan\left ( \frac{\pi}{4}+x \right )}{\tan\left ( \frac{\pi}{4}-x \right )}$
= $\LARGE \frac{\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}}{\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}}$
= $\LARGE \frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}$
= $\frac{1+\tan x}{1-\tan x}*\frac{1+\tan x}{1-\tan x}$
= $\left (\frac{1+\tan x}{1-\tan x} \right )^{2}$
L.H.S.=R.H.S.
8. $\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)}=\cot^{2}x$
L.H.S.,
= $\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$
= $\frac{\cos^{2}x}{\sin^{2}x}=\cot^{2}x$
L.H.S.=R.H.S.
9. $\cos \left ( \frac{3\pi}{2}+x \right )$ $\cos \left ( 2\pi+x \right )$ [$\cot \left ( \frac{3\pi}{2}+x \right )$+ $\cot \left ( 2\pi+x \right )$]=1
L.H.S.,
= sin x cos x [tan x + cot x]
= $\sin x\cos x\left [ \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \right ]$
= $\sin x\cos x\left [ \frac{\sin^{2} x+\cos^{2}x}{\cos x\sin x}\right ]$
= $\sin^{2} x+\cos^{2}x$ =1
L.H.S.=R.H.S.
10. sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x = cos x
L.H.S.,
= cos[(n+1)x-(n+2)x] [cos(x-y)=cos x cos y+sin x sin y]
= cos(n+1-n-2)x
= cos (-x) = cos x
L.H.S.=R.H.S.
11. $\cos \left ( \frac{3\pi}{4}+x \right )$-$\cos \left ( \frac{3\pi}{4}-x \right )$$=-\sqrt{2}\sin x$
We know that,
cos(x+y)=cos x cos y-sin x sin y
cos(x-y)=cos x cos y+sin x sin y
L.H.S.,
=$\cos \frac{3\pi}{4}\cos x$ $-\sin \frac{3\pi}{4}\sin x$ $-\cos \frac{3\pi}{4}\cos x$$-\sin \frac{3\pi}{4}\sin x$
= $-2\sin \frac{3\pi}{4}\sin x$
= $-2\sin \left ( \pi-\frac{\pi}{4} \right )\sin x$
= $-2\sin \left ( \frac{\pi}{4} \right )\sin x$
= $-2\left ( \frac{1}{\sqrt{2}} \right )\sin x$ = $-\sqrt{2}\sin x$
L.H.S.=R.H.S.
12. $\sin^{2}6x-\sin^{2}4x=\sin 2x \sin 10x$
We know that,
$\sin x+\sin y=2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\sin x-\sin y=2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
L.H.S.,
= (sin 6x-sin 4x)(sin 6x+sin 4x)
= $\left [ 2\cos\left ( \frac{6x+4x}{2} \right )\sin\left ( \frac{6x-4x}{2} \right ) \right ]$ $\left [ 2\sin\left ( \frac{6x+4x}{2} \right )\cos\left ( \frac{6x-4x}{2} \right ) \right ]$
= (2 cos 5x sin x)(2 sin 5x cos x)
= (2 cos 5x sin 5x) (2 cos x sin x)
= cos 10x sin 2x [sin 2x=2 sin x cos x]
L.H.S.=R.H.S.
13. $\cos^{2}2x-\cos^{2}6x=\sin 4x\sin 8x$
We know that,
$\cos x+\cos y=2\cos\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\cos x-\cos y=-2\sin\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
L.H.S.,
= (cos 2x-cos 6x)(cos 2x+cos 6x)
=$\left [ -2\sin\left ( \frac{2x+6x}{2} \right )\sin\left ( \frac{2x-6x}{2} \right ) \right ]$ $\left [ 2\cos\left ( \frac{2x+6x}{2} \right )\cos\left ( \frac{2x-6x}{2} \right ) \right ]$
= [-2 sin 4x sin(-2x)][2 cos 4x cos(-2x)]
= (2 sin 4x sin 2x) (2 cos 4x cos 2x)
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x [sin 2x=2 sin x cos x]
L.H.S.=R.H.S.
14. $\sin 2x+2\sin 4x$ $+\sin 6x$ $=4\cos^{2}x\sin 4x$
L.H.S.,
= [sin 2x + sin 6x]+2sin 4x
= $\left [ 2\sin\left ( \frac{2x+6x}{2} \right )\cos\left ( \frac{2x-6x}{2} \right ) \right ]+2\sin 4x$ [$\sin x+\sin y=2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$]
= [2 sin 4x cos(-2x)]+2sin 4x
= 2 sin 4x cos 2x + 2sin 4x
= 2 sin 4x [cos 2x + 1]
= 2 sin 4x [2cos$\displaystyle \small ^{2}$ x-1+1] [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
= 2 sin 4x (2cos$\displaystyle \small ^{2}$ x) = 4 cos$\displaystyle \small ^{2}$ x sin 4x
L.H.S.=R.H.S.
15. $\cot 4x\left ( \sin 5x+\sin 3x \right )$ $=\cot x\left ( \sin 5x-\sin 3x \right )$
L.H.S.,
=$\frac{\cos 4x}{\sin 4x}\left [ 2\sin\left ( \frac{5x+3x}{2} \right )\cos\left ( \frac{5x-3x}{2} \right ) \right ]$ [$\sin x+\sin y=2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$]
= $\frac{\cos 4x}{\sin 4x}\left [ 2\sin 4x\cos x \right ]$
= $2\cos 4x \cos x$
R.H.S.,
= $\frac{\cos x}{\sin x}\left [ 2\cos\left ( \frac{5x+3x}{2} \right )\sin\left ( \frac{5x-3x}{2} \right ) \right ]$ [$\sin x-\sin y=2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$]
= $\frac{\cos x}{\sin x}\left [ 2\cos 4x\sin x \right ]$
= $2\cos 4x \cos x$
L.H.S.=R.H.S.
16. $\large \frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\frac{\sin 2x}{\cos 10x}$
L.H.S.,
= $\large \frac{-2\sin\left ( \frac{9x+5x}{2} \right )\sin\left ( \frac{9x-5x}{2} \right )}{2\cos\left ( \frac{17x+3x}{2} \right )\sin\left ( \frac{17x-3x}{2} \right )}$
= $\frac{-2\sin 7x \sin 2x}{2 \cos 10x\sin 7x}$
= $-\frac{ \sin 2x}{ \cos 10x}$
L.H.S.=R.H.S.
17. $\large \frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$
L.H.S.,
= $\large \frac{2\sin\left ( \frac{5x+3x}{2} \right )\cos\left ( \frac{5x-3x}{2} \right )}{2\cos\left ( \frac{5x+3x}{2} \right )\cos\left ( \frac{5x-3x}{2} \right )}$
= $\frac{2\sin 4x \cos x}{2 \cos 4x\cos x}$
= $\tan 4x$
L.H.S.=R.H.S.
18. $\large \frac{\sin x-\sin y}{\cos x+\cos y}=\tan\left ( \frac{x-y}{2} \right )$
L.H.S.,
= $\large \frac{2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )}{2\cos\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )}$
= $\tan\left ( \frac{x-y}{2} \right )$
L.H.S.=R.H.S.
19. $\large \frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 4x$
L.H.S.,
= $\large \frac{2\sin\left ( \frac{x+3x}{2} \right )\cos\left ( \frac{x-3x}{2} \right )}{2\cos\left ( \frac{x+3x}{2} \right )\cos\left ( \frac{x-3x}{2} \right )}$
= $\frac{\sin 2x }{ \cos 2x}$
= $\tan 2x$
L.H.S.=R.H.S.
20. $\large \frac{\sin x-\sin 3x}{\sin^{2}x-\cos^{2}x}=2\sin x$
L.H.S.,
= $\large \frac{2\cos\left ( \frac{x+3x}{2} \right )\sin\left ( \frac{x-3x}{2} \right )}{-\cos 2x}$
= $\large -\frac{2 \cos 2x \sin (-x)}{\cos 2x}$
= $\large 2\sin x$
L.H.S.=R.H.S.
21. $\large \frac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$
L.H.S.,
= $\frac{\left (\cos 4x+\cos 2x \right )+\cos 3x}{\left (\sin 4x+\sin 2x \right )+\sin 3x}$
= $\large \frac{2\cos\left ( \frac{4x+2x}{2} \right )\cos\left ( \frac{4x-2x}{2} \right )+\cos 3x}{2\sin\left ( \frac{4x+2x}{2} \right )\cos\left ( \frac{4x-2x}{2} \right )+\sin 3x}$
= $\frac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x}$
= $\frac{\cos 3x(2\cos x+1)}{\sin 3x(2\cos x+1)}$
= $\cot 3x$
L.H.S.=R.H.S.
22. $\cot x\cot 2x$ $-\cot 2x\cot 3x$ $-\cot 3x\cot x=1$
L.H.S.,
= $\cot x\cot 2x-\cot 3x(\cot 2x+\cot x)$
= $\cot x\cot 2x$ $-\cot (2x+x)(\cot 2x+\cot x)$
= $\cot x\cot 2x$ $-\left [ \frac{\cot 2x\cot x-1}{\cot 2x+\cot x} \right ]$ $(\cot 2x+\cot x)$ [$\cot(x+y)=\frac{\cot x\cot y-1}{\cot x+\cot y}$ ]
= $\cot x\cot 2x-\cot 2x\cot x+1$ = 1
L.H.S.=R.H.S.
23. $\tan 4x=\frac{4\tan x(1-\tan^{2}x)}{1-6\tan^{2}x+\tan^{4}x}$
L.H.S.,
= $\tan 2(2x)$
= $\frac{2\tan 2x}{1-\tan^{2}2x}$ [$\tan 2x=\frac{2\tan x}{1-\tan^{2}x}$ ]
= $\large \frac{2\left (\frac{2\tan x}{1-\tan^{2}x} \right )}{1-\left (\frac{2\tan x}{1-\tan^{2}x} \right )^{2}}$
= $\large \frac{2\left (\frac{2\tan x}{1-\tan^{2}x} \right )}{1-\frac{4\tan^{2}x}{(1-\tan^{2}x)^{2}}}$
= $\large \frac{2\left (\frac{2\tan x}{1-\tan^{2}x} \right )}{\frac{(1-\tan^{2}x)^{2}-4\tan^{2}x}{(1-\tan^{2}x)^{2}}}$
= $\frac{4\tan x}{1-\tan^{2}x}*\frac{(1-\tan^{2}x)^{2}}{(1-\tan^{2}x)^{2}-4\tan^{2}x}$
= $\frac{4\tan x(1-\tan^{2}x)}{1+\tan^{4}x-2\tan^{2}x-4\tan^{2}x}$
= $\frac{4\tan x(1-\tan^{2}x)}{1+\tan^{4}x-6\tan^{2}x}$
L.H.S.=R.H.S.
24. $\cos 4x=1-8\sin^{2}x\cos^{2}x$
L.H.S.,
= $\cos 2(2x)$
= $1-2\sin^{2}2x$ [cos 2x=1-2sin$\displaystyle \small ^{2}$ x]
= $1-2(2\sin x\cos x)^{2}$
= $1-8\sin^{2} x\cos^{2} x$
L.H.S.=R.H.S.
25. $\cos 6x=32\cos^{6}x$ $-48\cos^{4}x+18\cos^{2}x-1$
L.H.S.,
= $\cos 3(2x)$
= $4\cos^{3}2x-3\cos 2x$ [cos 3x=4cos$\displaystyle \small ^{3}$ x-3cosx]
= $4[2\cos^{2}x-1]^{3}-3[2\cos^{2}x-1]$
= $4[(2\cos^{2}x)^{3}-1^{3}-3(2\cos^{2}x)^{2}$ $+3(2\cos^{2}x)]-6\cos^{2}x+3$
= $32\cos^{6}x-48\cos^{4}x+24\cos^{2}x$ $-4-6\cos^{2}x+3$
= $32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1$
L.H.S.=R.H.S.
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