\(\text { Let } \theta=\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right), \theta \in \text { Ist quadrant }\left([0, \pi]-\left\{\frac{\pi}{2}\right\}\right) \)
\(\Rightarrow \sec \theta=\frac{2}{\sqrt{3}} \Rightarrow \sec \theta=\sec \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6}\)

\(\frac { \pi }{ 3 } \)

\({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) =\frac { \pi }{ 6 } \)
Alternative Method :
\({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) ={ tan }^{ -1 }\left[ tan\left( \pi +\frac { \pi }{ 6 } \right) \right] \)
\(={ tan }^{ -1 }\left[ tan\frac { \pi }{ 6 } \right] =\frac { \pi }{ 6 } \)
\(\left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)

\(tan^{ -1 }\left( tan\frac { 2\pi }{ 3 } \right) ={ tan }^{ -1 }\left( -tan\frac { \pi }{ 3 } \right) \)
\(=-\frac { \pi }{ 3 } \)

cot (tan^-1a + cot^-1a) = 0
Alternative Method :
\(cot\ (tan^{ -1 }a+{ cot }^{ -1 }a)=cot\left( \frac { \pi }{ 2 } \right) \)
\(\left[ \because { tan }^{ -1 }x+{ cot }^{ -1 }x=\frac { \pi }{ 2 } \right] \)
= 0

\( \sin { \left( 2\cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } \)
\(=2\sin { \left( 2\cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } cos\left( { cos }^{ -1 }\left( -\frac { 3 }{ 5 } \right) \right) \)
\(\\ =2\sin { \left( \cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } \left( -\frac { 3 }{ 5 } \right) \)
\(=-2\sin { \left( \sin ^{ -1 }{ \left( \frac { 4 }{ 5 } \right) } \right) } \left( -\frac { 3 }{ 5 } \right) \)
\(=-2\left( \frac { 4 }{ 5 } \right) \left( -\frac { 3 }{ 5 } \right) \)
\(=+\frac { 24 }{ 25 } \)

\(\sqrt { ab } \)

\(-sin^{ -1 }\left( \frac { x }{ a } \right) \)

\( \mathrm{LHS}=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \\ \text { Put } x=\sin \theta, \text { then } \theta=\sin ^{-1} x \\ \therefore \mathrm{LHS}=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right)=\sin ^{-1}(2 \sin \theta \cos \theta) \\ \left[\because \sin ^{2} \theta+\cos ^{2} \theta=1 \Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}\right]\\ =\sin ^{-1}(\sin 2 \theta) [\because 2 \sin A \cos A=\sin 2 A] \\ =2 \theta=2 \sin ^{-1} x \quad\left[\because x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\right] \)