Let \(\cos^{-1}(\frac12)=\theta\)
\(\Rightarrow\frac12=\cos\theta\Rightarrow\cos=\frac{\pi}{3}\cos\theta\)
\(\Rightarrow\theta=\frac{\pi}{3}\)
\(\therefore\cos\left[\frac{\pi}{3}-\cos^{-1}(\frac{1}{2})\right]=\cos\left[\frac{\pi}{3}-\frac{\pi}{3}\right]=\cos(0)=1.\)

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.

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 = \(\left( \frac { 3 }{ 5 } \right) \left( -\frac { 12 }{ 13 } \right) -\left( \frac { 4 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) \)
= \(\frac { -36 }{ 65 } +\frac { 20 }{ 65 } =\frac { -16 }{ 65 } \)

\(\sin { { 300 }^{ o } } \)
\(=\sin { 300 }^{ o } =\sin { \left( { 360 }^{ o }-{ 60 }^{ o } \right) } \) (IV quad)
\(= \sin 60^o=-\frac { \sqrt { 3 } }{ 2 } \)

\({ 150 }^{ o }=150\times \frac { \pi }{ 180 } = \frac { 5\pi }{ 6 } \)

LHS \(=\tan^{-1}\left(\frac{4}{3}\right)-\tan^{-1}\left(\frac{1}{7}\right)\)
\(=\tan^{-1}\left(\frac{\frac{4}{3}-\frac{1}{7}}{1+\frac{4}{3}.\frac{1}{7}}\right)\)
\(=\tan^{-1}\left(\frac{\frac{28 - 3}{21}}{\frac{21+4}{21}}\right)\)
\(=\tan^{-1}(\frac{25}{25})=\tan^{-1}(1)=\frac{\pi}{4}=RHS\)
Hence proved.

\(\cos (\alpha+\beta)=\frac{12}{13} \Rightarrow \sin (\alpha+\beta)=\frac{5}{13} \)
\(\sin (\alpha-\beta)=3 / 5 \Rightarrow \cos (\alpha-\beta)=4 / 5 \)
\(\sin 2 \alpha=\sin ((\alpha+\beta)+(\alpha-\beta)) \)
\(= \sin (\alpha+\beta) \cos (\alpha-\beta) +\cos (\alpha+\beta) \sin (\alpha-\beta) \)
\(= \frac{5}{13} \cdot \frac{4}{5}+\frac{3}{5} \cdot \frac{12}{13}=\frac{56}{65}\)

\(\frac{3 \tan 10^{\circ}-\tan ^3 10^{\circ}}{1-3 \tan ^2 10^{\circ}} =\tan 3(10) =\tan 30^{\circ}=\frac{1}{\sqrt{3}} \)

cos²45^o - sin²45^o = cos 2(45^o)
= cos 90^o = 0

\(\sin \left(-420^{\circ}\right) =-\sin 420^{\circ}=-\sin (360+60) =-\sin 60^{\circ}=\frac{-\sqrt{3}}{2} \)

\(37^{\circ} 30^{\prime}=37 \frac{1}{2}^{\circ}=\frac{75}{2} \times \frac{\pi}{180}=\frac{5 \pi}{24}\)

\(\tan\left( x+y \right) =42\)
\(x+y={ \tan }^{ -1 }\left( 42 \right) \)
\({ \tan }^{ -1 }\left( 2 \right) +y={ \tan }^{ -1 }(42)\)
\(y={ \tan }^{ -1 }(42)-{ \tan }^{ -1 }(2)\)
\(={ \tan }^{ -1 }\left[ \cfrac { 42-2 }{ 1+\left( 42\times 2 \right) } \right] \)
\(={ \tan }^{ -1 }\left( \cfrac { 40 }{ 85 } \right) \)
\( y={ \tan }^{ -1 }\left[ \cfrac { 8 }{ 17 } \right] \)

LHS\(=\frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}\)
\(\left[\because\sin C\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)\ and\ \cos C\cos D=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)\right]\)
\(=\frac{2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)}{{2\cos\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)}}=\frac{2\sin4x.\cos x}{2\cos4x.\cos x}=\tan4x=RHS\)
Hence proved.

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.

\(\cos { { 510 }^{ o } } =\cos { \left( { 360 }^{ o }+{ 150 }^{ o } \right) } =\cos { { 150 }^{ o } } =\cos { \left( { 180 }^{ o }-{ 30 }^{ o } \right) } =-\cos { { 30 }^{ o } } =-\frac { \sqrt { 3 } }{ 2 } \)
\(\cos { { 330 }^{ o } } =\cos { \left( { 360 }^{ o }-{ 30 }^{ o } \right) } =\cos { { 30 }^{ o } } =\frac { \sqrt { 3 } }{ 2 } \)
\(\sin { { 390 }^{ o } } =\sin { \left( { 360 }^{ o }+{ 30 }^{ o } \right) } =\sin { { 30 }^{ o } } =\frac { 1 }{ 2 } \)
\(\cos { { 120 }^{ o } } =\cos { \left( { 90 }^{ o }+{ 30 }^{ o } \right) } =-\cos { { 60 }^{ o } } =-\frac { 1 }{ 2 } \)
\(\therefore LHS=\cos { { 510 }^{ o } } \cos { { 330 }^{ o } } +\sin { { 390 }^{ o } } \cos { { 120 }^{ o } } \)
\(=\left( -\frac { \sqrt { 3 } }{ 2 } \right) \left( \frac { \sqrt { 3 } }{ 2 } \right) +\left( \frac { 1 }{ 2 } \right) \left( \frac { -1 }{ 2 } \right) =-\frac { 3 }{ 4 } -\frac { 1 }{ 4 } =\frac { -4 }{ 4 } =-1\)
= RHS.
Hence Proved.

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

LHS = cos 4x = cos 2(2x)
= 2 cos²2x-1 [∴ cos 2A = 2cos²A - 1]
= 2 [2 cos²x - 1]² - 1 = 2 [4 cos⁴x - 4cos2x + 1] - 1
= 8 cos⁴x - 8 cos²x + 2 - 1 = 1 - 8 cos²x + 8 cos⁴x
= 1 - 8 cos²x (1 - cos²x) = 1 - 8 cos²x . sin²x = RHS.

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

Since \(\alpha+\beta\) and \(\alpha-\beta\) are acute
\(\cos (\alpha-\beta)\) and \(\sin (\alpha+\beta)\) are positive
\(\sin (\alpha+\beta)=\sqrt{1-\cos ^2(\alpha+\beta)}\)
\(=\sqrt{1-\frac{16}{25}}\)
\(=\sqrt{\frac{9}{25}}=\frac{3}{5}\)
\(\cos (\alpha-\beta)=\sqrt{1-\sin ^2(\alpha-\beta)}\)
\(=\sqrt{1-\frac{25}{169}}\)
\(=\sqrt{\frac{144}{169}}\)
\(=\frac{12}{13}\)
\(\tan (\alpha+\beta)=\frac{\sin (\alpha+\beta)}{\cos (\alpha+\beta)}\)
\(=\frac{3 / 5}{4 / 5}=\frac{3}{4}\)
\(\tan (\alpha-\beta)=\frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}\)
\(=\frac{5 / 13}{12 / 13}=\frac{5}{12}\)
\(\tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)]\)
\(=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)}\)
\(=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}\)
\(=\frac{36+20}{48-15}\)
\(=\frac{56}{33}\)