Using \(\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)\), we get
\(LHS=\sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right)=\sin\left(\frac{\pi}{8}+\frac x2+\frac{\pi}{8}-\frac x2\right).\sin\left(\frac{\pi}{8}+\frac x2-\frac{\pi}{8}+\frac x2\right)\)
\(=\sin\left(\frac{2\pi}{8}\right).\sin\left(\frac{2x}{2}\right)=\sin\left(\frac{\pi}{4}\right).\sin x=\frac{1}{\sqrt2}.\sin x=RHS\)
Hence proved.

\(\cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{4}+\frac{\pi}{6}\right)\)
\(=\cos\frac{\pi}{4}\cos\frac{\pi}{6}-\sin\frac{\pi}{4}\sin\frac{\pi}{6}[\therefore \cos(A+B)=\cos A\cos B-\sin A\sin B]\)
\(=\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times\frac{1}{2}=\frac{\sqrt3-1}{2\sqrt2}\)

A lies in I quadrant and B lies in the IlI quadrant

\(\cos A=\sqrt{1-\sin ^2 A}\)
\(=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\cos B=-\frac{12}{13}\)
\(\sin B=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=\frac{-5}{13}\)
sin (A + B)
 = sin A cos B - cos A sin B = \(\left( \frac { 3 }{ 5 } \right) \left( -\frac { 12 }{ 13 } \right) -\left( \frac { 4 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) \)
= \(\frac { -36 }{ 65 } +\frac { 20 }{ 65 } =\frac { -16 }{ 65 } \)

-140° III quadrant.

Let cosec^-1(2) = y
\(\sin ^{-1}\left(\frac{1}{2}\right)=y\) where \(\frac{-\pi}{2}\le y\le \frac{\pi}{2}\)
\(\sin y=1 / 2=\sin (\pi / 6)\)
\(\Rightarrow y=\frac{\pi}{6}\)

Let \(\sin^{-1}\left(\frac{-1}{2}\right)\)=y where \(\frac{-\pi}2\le y\le\frac{\pi}{2}\)
\(\therefore \sin y=-\frac{1}{2}-\sin\left(\frac{-\pi}{6}\right)\)
\(\Rightarrow y=\frac{-\pi}{6}\)

\((\sin A+\cos A)^2=1\)
\(\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}+2 \sin \mathrm{A} \cos \mathrm{A}=1\)
\(1+\sin 2 \mathrm{~A}=1 \Rightarrow \sin 2 \mathrm{~A}=0\)

\(\sec A(-\cos A)=\frac{1}{\cos A}(-\cos A)=-1\)

\(\frac{1}{2}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) =\frac{1}{2} \sin 2\left(15^{\circ}\right)=\frac{1}{2} \sin 30^{\circ} =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)

\(\sec \theta =\sqrt{1+\tan ^2 \theta}=\sqrt{1+\frac{1}{5}}=\sqrt{\frac{6}{5}} \)

\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=22 \frac{1}{2}^{\circ}=22^{\circ} 30^{\prime}\)

LHS\(=\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)\)
\(=\frac{2\sin\left(\frac{\pi}{3}+x\right).\sin\left(\frac{\pi}{3}-x\right)}{2\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)}\)
\([\because2\sin A\sin B=\cos(A-B)-\cos(A+B)\ and\ \ 2\cos A\cos B=\cos(A+B)+\cos(A-B)]\)
\(=\frac{\cos\left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}+x+\frac{\pi}{3}-x\right)}{\cos\left(\frac{\pi}{3}+x+\frac{\pi}{3}-x\right)+\cos\left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right)}\)
\(=\frac{\cos2x-\cos\frac{2\pi}{3}}{\cos\frac{2\pi}{3}+\cos2x}=\frac{\cos2x+\frac{1}{2}}{-\frac12+\cos2x}\)
\(=\frac{2\cos2x+1}{2\cos2x-1}\)
=RHS
Hence proved.

Consider \(\tan(\alpha+\beta)\)
\(=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\left(\frac{m}{m+1}\right)\left(\frac{1}{2m+1}\right)}=\frac{\frac{m(2m+1)+1(m+1)}{(m+1)(2m+1)}}{\frac{(m+1)(2m+1)-m}{(m+1)(2m+1)}}\)
\(=\frac{\frac{2m^2+m+m1}{(m+1)(2m+1)}}{\frac{2m^2+m+2m+1-m}{(m+1)(2m+1)}}=\frac{2m^2+2m+1}{(m+1)(2m+1)}\times\frac{(m+1)(2m+1)}{2m^2+2m+1}\)
\(=1=\tan\frac{\pi}{4}\therefore\tan(\alpha+\beta)=\tan\frac{\pi}{4}\Rightarrow\alpha+\beta=\frac{\pi}{4}\)

\(2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2} \)
\(\operatorname{cosec}^2 \frac{7 \pi}{6} =\operatorname{cosec}^2\left(\pi+\frac{\pi}{6}\right) \)
\(=\operatorname{cosec}^2 \frac{\pi}{6}=\operatorname{cosec}^2 30^{\circ}=2 \)
\(\text { LHS }= 2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{\pi}{6} \cos ^2 \frac{\pi}{3} \)
\(= 2 \sin ^2 30^{\circ}+2^2\left(\cos ^2 60^{\circ}\right) \)
\(= 2\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)^2 \)
\(= \frac{2}{4}+\frac{4}{4}=\frac{6}{4}=3 / 2=\mathrm{RHS} \)
Hence proved.

LHS \(=\sqrt{3}\ cosec{20^o}-\sec20^o=\sqrt{3}.\frac{1}{\sin20^o}-\frac{1}{\cos20^o}\)
\(=\frac{\sqrt3\cos20^o-\sin20^o}{\sin20^o\cos20^o}=2\left[\frac{\frac{\sqrt3}{2}\cos20^o-\frac{1}{2}\sin20^o}{\sin20^o\cos20^o}\right]\)
\(=2\frac{(\sin60^o\cos20^o-\cos60^o\sin20^o)}{\sin20^o\cos20^o}\left[\because\sin60^o=\frac{\sqrt3}{2}\ and\cos60^o=\frac{1}{2}\right]\)
\(=2\frac{\sin(60^o-20^o)}{\sin20^o\cos20^o}=\frac{4\sin40^o}{2\sin20^o\cos20^o}=\frac{4\sin40^o}{\sin40^o}[\because2\sin A\cos A=\sin2A]\)
= 4 = RHS
Hence proved

LHS = cot x cot 2x - cot 2 x cot 3 x - cot 3x cot x
We have 3x = x + 2x
\(cot\quad 3x=\frac { cotx\quad cot2x-1 }{ cotx+cot2x } \)   \(\left[ \therefore tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB } \right] \)
cross multiplying we get,
cot x cot 3x + cot 2x cot 3x = -1 + cot x cot 2x
\(\Rightarrow\) cot x cot 2x - cot 2x cot 3x - cot x cot 3x = 1
Hence proved

LHS = cos²2x - cos²6x
= (cos 2x + cos 6x)(cos 2x - cos 6x) [∴ cos²A - cos²B = (cos A + cos B)(cos A - cos B)
\(=\left[ 2cos\left( \frac { 2x+6x }{ 2 } \right) .cos\left( \frac { 2x-6x }{ 2 } \right) \right] \left[ -2sin\left( \frac { 2x+6x }{ 2 } \right) .sin\left( \frac { 2x-6x }{ 2 } \right) \right] \)
\(=\left[ \because cosC+cosD=2\quad cos\left( \frac { C+D }{ 2 } \right) cos\left( \frac { C-D }{ 2 } \right) and\quad cosC-cosD=-2sin\left( \frac { C+D }{ 2 } \right) .sin\left( \frac { C-D }{ 2 } \right) \right] \)
= [2 cos 4x . cos(-2x)][-2 sin 4x . sin(-2x)]
= (2 cos 4x . cos 2x) (2 sin 4x . sin 2x) [:. cos (-\(\theta\)) = cos \(\theta\) and sin (-\(\theta\)) = -sin\(\theta\)]
= (2 sin 2x cos 2x) (2 sin 4x cos 4x)
= sin 4x sin 8x [sin 2A = 2 sin A cos A] = RHS.
Hence proved

\(\frac { \sin { \left( { 180 }^{ o }+A \right) \cos { \left( { 90 }^{ o }-A \right) \tan { \left( { 270 }^{ o }-A \right) } } } \quad \quad }{ \sec { \left( { 540 }^{ o }-A \right) \cos { \left( { 360 }^{ o }+A \right) \csc { \left( { 270 }^{ o }+A \right) } } } } \)
\(=\frac{(-\sin A)(\sin A)(\cot A)}{\sec (360+180-A) \cos A(-\sec A)} \)
\(=\frac{(-\sin A)(\sin A) \frac{\cos A}{\sin A}}{(-\sec A) \cos A(-\sec A)} \)
\(=-\frac{\sin A \cos A}{\frac{1}{\cos A} \cos A \frac{1}{\cos A}} \)
\(=-\sin A \cos ^2 A\)
= R.H.S
Hence proved.

\(\text { LHS }=\sin \theta \cdot \cos \theta\left\{\sin \left(\frac{\pi}{2}-\theta\right) \cdot \operatorname{cosec} \theta\right. \left.+\cos \left(\frac{\pi}{2}-\theta\right) \cdot \sec \theta\right\} \)
\(=\sin \theta \cos \theta\left\{\cos \theta\left(\frac{1}{\sin \theta}\right)+\sin \theta\left(\frac{1}{\cos \theta}\right)\right\} \)
\(=\cos ^2 \theta+\sin ^2 \theta=1=\text { RHS } \)
Hence proved.