\(\text { Let } \theta=\tan -1 \sqrt{3} \)
\(\Rightarrow \tan \theta=\sqrt{3} \)
\(\Rightarrow \tan \theta=\tan \frac{\pi}{3} \)
\(\Rightarrow \theta=\frac{\pi}{3} \)
\(\therefore \quad \tan ^{-1} \sqrt{3}=\frac{\pi}{3} \)
\(\text { Again let } \theta=\sec ^{-1}(-2) \)
\(\Rightarrow \sec \theta=\sec \frac{2 \pi}{3} \)
\(=\sec \left(\pi-\frac{\pi}{3}\right) \)
\(\Rightarrow \sec \theta=\sec \frac{2 \pi}{3} \)
\(\Rightarrow \theta=\frac{2 \pi}{3}\)
\(\Rightarrow \sec ^{-1}(-2)=\frac{2 \pi}{3}\)
\(\text { From ( } i \text { ) and (ii), we get }\)
\(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3} \)

\(\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } -\tan ^{ -1 }{ \left( \frac { a-b }{ a+b } \right) } =\frac { \pi }{ 4 } \)
Alternative Method :
\(\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } -\tan ^{ -1 }{ \left( \frac { a-b }{ a+b } \right) } =\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } -\tan ^{ -1 }{ \left( \frac { \frac { a }{ b } -1 }{ \frac { a }{ b } +1 } \right) } \)
By taking \(\frac { a }{ b } =tan\quad \theta \)
\({ tan }^{ -1 }(tan\quad \theta )-{ tan }^{ -1 }\left( \frac { tan\quad \theta -tan\left( \frac { \pi }{ 4 } \right) }{ 1+tan\quad \theta \quad tan\left( \frac { \pi }{ 4 } \right) } \right) \)
\(={ tan }^{ -1 }(tan\quad \theta )-{ tan }^{ -1 }\left[ tan\left( \theta -\frac { \pi }{ 4 } \right) \right] \)
\(=\theta -\theta +\frac { \pi }{ 4 } \ \left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
\(=\frac { \pi }{ 4 } \)

\(sin\left( 2sin^{ -1 }\frac { 3 }{ 5 } \right) =\frac { 24 }{ 25 } \)
Alternative Method :
\(sin\left( 2sin^{ -1 }\frac { 3 }{ 5 } \right) =sin\left( 2tan^{ -1 }\frac { 3 }{ 4 } \right) \)
\(=sin\left[ { tan }^{ -1 }\left( \frac { 2\times \frac { 3 }{ 4 } }{ 1-\frac { 9 }{ 16 } } \right) \right] \)
\(=sin\left[ { tan }^{ -1 }\left( \frac { 3 }{ 2 } \times \frac { 16 }{ 7 } \right) \right] \)
\(=sin\left[ { tan }^{ -1 }\frac { 24 }{ 7 } \right] \)
\(=sin\left[ { sin }^{ -1 }\frac { 24 }{ 25 } \right] \)
\(=\frac { 24 }{ 25 } \)
\(\\ \left[ \because { sin }^{ -1 }(sin\quad \theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)