\(\text { We have, } \tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\tan ^{-1}\left[\frac{\sin \left(\frac{\pi}{2}-x\right)}{1-\cos \left(\frac{\pi}{2}-x\right)}\right]\)
\(=\tan ^{-1}\left[\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right] \)
\(=\tan ^{-1} \cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\)
\(=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{x}{2}\right)\right)\right]\)
\(\text { Now, }-\frac{3 \pi}{2}<x<\frac{\pi}{2}\)
\(\Rightarrow \quad-\frac{3 \pi}{4}<\frac{x}{2}<\frac{\pi}{4} \Rightarrow-\frac{\pi}{2}<\frac{x}{2}+\frac{\pi}{4}<\frac{\pi}{2}\)
\(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\tan ^{-1} \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\pi}{4}+\frac{x}{2} \)
\({\left[\because \tan \left(\tan ^{-1} x\right)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]}\)

\( \sin ^{-1}\left[\sin \left(\frac{13 \pi}{7}\right)\right]=\sin ^{-1}\left[\sin \left(2 \pi-\frac{\pi}{7}\right)\right] =\sin ^{-1}\left[\sin \left(\frac{-\pi}{7}\right)\right]=\frac{-\pi}{7} \quad\left[\because \frac{-\pi}{7} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\right] \)

\(f(x)=\sin ^{-1} x, x \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]\)

\(3 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}(0) \)
\(=3 \times \frac{\pi}{4}+2 \times \frac{\pi}{6}+\frac{\pi}{2} \)
\( =\frac{3 \pi}{4}+\frac{\pi}{3}+\frac{\pi}{2} \)
\( =\frac{9 \pi+4 \pi+6 \pi}{12}=\frac{19 \pi}{12}\)

We have, \(y=\sin ^{-1}\left(x^2-4\right)\)
\(\Rightarrow  -1 \leq x^2-4 \leq 1\)
\(\Rightarrow -1+4 \leq x^2 \leq 1+4\)
\(\Rightarrow  3 \leq x^2 \leq 5\)
\(\Rightarrow  \sqrt{3} \leq|x| \leq \sqrt{5}\)
\(\Rightarrow  x \in[-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]\)
Therefore The domain of y is \([-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]\)

We know that domain and range (principal value branch) of \(f(x)=\tan ^{-1} x\) are R and \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) respectively.

\(\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})= \tan ^{-1} \sqrt{3}-\left(\pi-\cot ^{-1} \sqrt{3}\right)\)
\(\left[\because \cot ^{-1}(-x)=\pi-\cot ^{-1}(x), x \in R\right] \)
\(= \tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}-\pi \)
\(= \frac{\pi}{2}-\pi \quad\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, x \in R\right] \)
\(=-\frac{\pi}{2}\)

\(2 tan-1(cos x) = tan-1(2 cosec x) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2cosx }{ 1-{ cos }^{ 2 }x } \right) ={ tan }^{ -1 }(2\quad cosec\quad x) \)
\(\Rightarrow 2 cot x cosec x = 2 cosec x \)
\(\Rightarrow cot x = 1\)
\(\Rightarrow x={ cot }^{ -1 }(1)=\frac { \pi }{ 4 }\)

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

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

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

\(\cot \left(\tan ^{-1} a+\cos ^{-1}\right) \)
\(=\cot \left(\frac{\pi}{2}\right) \quad\left[\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \)
\(=0\)

\(\left[ { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) -2{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right] =\frac { 2\pi }{ 3 } \)
Alternative Method :
\(\left[ { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) -2{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right] \)
\(=\left[ { cos }^{ -1 }\left( cos\frac { \pi }{ 3 } \right) -2sin^{ -1 }\left( -sin\frac { \pi }{ 6 } \right) \right] \)
\(=\left[ \frac { \pi }{ 3 } -2\times \left( -\frac { \pi }{ 6 } \right) \right] \)
\([\because { cos }^{ -1 }(cos\quad \theta )=\theta \forall \theta \in [0,\quad \pi ]\quad and\quad { sin }^{ -1 }(sin\quad \theta )=\theta \forall \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] ]\)
\(=\left[ \frac { \pi }{ 3 } +\frac { \pi }{ 3 } \right] =\frac { 2\pi }{ 3 } \)

\(\left[\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\right]=\frac{2 \pi}{3}\)
Alternative Method:
\( {\left[\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\right]}\)
\( =\left[\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)\right]\)
\( =\left[\frac{\pi}{3}+2 \times \frac{\pi}{6}\right]\)
\( {\left[\because \quad \cos ^{-1}(\cos \quad \theta)=\theta \forall \theta[0, \quad \pi] \quad \text { and } \sin ^{-1}(\sin \quad \theta)=\theta \forall \theta=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]} \)
\( =\frac{\pi}{3}+\frac{\pi}{3} =\frac{2 \pi}{3}\)

\({ tan }^{ -1 }\sqrt { 3 } -{ sec }^{ -1 }(-2)=-\frac { \pi }{ 3 } \)
Alternative Method :
\({ tan }^{ -1 }\sqrt { 3 } -{ sec }^{ -1 }(-2)\)
\(=tan^{ -1 }\left[ tan\left( \frac { \pi }{ 3 } \right) \right] -{ sec }^{ -1 }\left[ -sec\left( \frac { \pi }{ 3 } \right) \right] \)
\(=\frac { \pi }{ 3 } -\left( \pi -\frac { \pi }{ 3 } \right) \)
\(\left[ \because \quad { tan }^{ -1 }(tan\quad \theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \quad and\quad { sec }^{ -1 }(sec\quad \theta )=\theta \forall \theta \in (0,\quad \pi )-\left( \frac { \pi }{ 2 } \right) \right] \)
\(=\frac { 2\pi }{ 3 } -\pi =-\frac { \pi }{ 3 } \)

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

\({ tan }^{ -1 }\left( tan\frac { 9\pi }{ 8 } \right) =\frac { \pi }{ 8 } \)
Alternative Method :
\({ tan }^{ -1 }\left( tan\frac { 9\pi }{ 8 } \right) ={ ta }n^{ -1 }\left[ tan\left( \pi +\frac { \pi }{ 8 } \right) \right] \)
\(={ tan }^{ -1 }\left[ tan\frac { \pi }{ 8 } \right] \)
\(\left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
\(=\frac { \pi }{ 8 } \)

\({ tan }^{ -1 }\left[ 2sin\left( 2{ cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \right) \right] =\frac { \pi }{ 3 } \)
Alternative Method :
\({ tan }^{ -1 }\left[ 2sin\left( 2{ cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \right) \right] \)
\(={ tan }^{ -1 }\left[ 2sin\left( 2{ cos }^{ -1 }\left( cos\frac { \pi }{ 6 } \right) \right) \right] \)
\(={ tan }^{ -1 }\left[ 2sin\left( 2\left( \frac { \pi }{ 6 } \right) \right) \right] ={ tan }^{ -1 }\left[ 2\times \frac { \sqrt { 3 } }{ 2 } \right] \)
\(\Rightarrow { tan }^{ -1 }\left( \sqrt { 3 } \right) =\frac { \pi }{ 3 } \)

\(\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)\right]=\frac{5}{12}\)
Alternative Method:
\( \tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)\right]=\tan \left[\tan ^{-1}\left(\frac{2\left(\frac{1}{5}\right)}{1-\frac{1}{25}}\right)\right]\)
\( =\tan \left[\tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right)\right] \)
\( {\left[\because \tan ^{-1}(\tan \theta)=\theta \forall \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right]}\)
\( =\left(\frac{5}{12}\right) \)

\({ tan }^{ -1 }(1)+{ cos }^{ -1 }\left( -\frac { 1 }{ 2 } \right) =\frac { 11\pi }{ 12 } \)
Alternative Method :
\({ tan }^{ -1 }(1)+{ cos }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \)
\(={ tan }^{ -1 }\left( tan\frac { \pi }{ 4 } \right) +{ cos }^{ -1 }\left( -cos\frac { \pi }{ 3 } \right) \)
\(={ tan }^{ -1 }\left( tan\frac { \pi }{ 4 } \right) +{ cos }^{ -1 }\left[ cos\left( \pi -\frac { \pi }{ 3 } \right) \right] \)
\(\left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \& { cos }^{ -1 }(cos\quad \theta )=\theta \forall \theta \in (0,\pi ) \right] \)
\(=\frac { \pi }{ 4 } +\frac { 2\pi }{ 3 } =\frac { 11\pi }{ 12 } \)

\({ sin }^{ -1 }\left[ sin\left( \frac { 3\pi }{ 5 } \right) \right] =\frac { 2\pi }{ 5 } \)
Alternative Method :
\({ sin }^{ -1 }\left[ sin\left( \frac { 3\pi }{ 5 } \right) \right] ={ sin }^{ -1 }\left[ sin\left( \pi -\frac { 3\pi }{ 5 } \right) \right] \)
\(\left( \because \quad \frac { 3\pi }{ 5 } \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right) \)
\({ sin }^{ -1 }\left( sin\frac { 2\pi }{ 5 } \right) =\frac { 2\pi }{ 5 } \)

\(\left[ { tan }^{ -1 }\left( -\sqrt { 3 } \right) +{ tan }^{ -1 }(1) \right] =-\frac { \pi }{ 12 } \)
Alternative Method :
\({ tan }^{ -1 }\left( -\sqrt { 3 } \right) +{ tan }^{ -1 }(1)\)
\(={ tan }^{ -1 }\left[ -tan\left( \frac { \pi }{ 3 } \right) \right] +{ tan }^{ -1 }\left( tan\frac { \pi }{ 4 } \right) \)
\(=-\frac { \pi }{ 3 } +\frac { \pi }{ 4 } =-\frac { \pi }{ 12 } \ \left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)

\(\left[ { cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) +{ cos }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right] \)\(=\frac { 5\pi }{ 6 } \)
Alternative Method
\(\left[ { cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) +{ cos }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right] \)
\(={ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 6 } \right) \right] +{ cos }^{ -1 }\left[ -cos\left( \frac { \pi }{ 3 } \right) \right] \)
\(={ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 6 } \right) \right] +{ cos }^{ -1 }\left[ cos\left( \pi -\frac { \pi }{ 3 } \right) \right] \)
\(=\frac { \pi }{ 6 } +\frac { 2\pi }{ 3 } =\frac { \pi }{ 6 } +\frac { 4\pi }{ 6 } \quad \left[ \because \quad { cos }^{ -1 }(cos\theta )=\theta \forall \theta \quad (0,\quad \pi ) \right] \)
\(=\frac { 5\pi }{ 6 } \)

\({ cos }^{ -1 }\left( -\frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) =\pi \)
Alternative Method :
\({ cos }^{ -1 }\left( -\frac { 1 }{ 2 } \right) +2{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(={ cos }^{ -1 }\left[ -cos\left( \frac { \pi }{ 3 } \right) \right] +2{ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 6 } \right) \right) \)
\(={ cos }^{ -1 }\left[ cos\left( \pi -\frac { \pi }{ 3 } \right) \right] +2\left( \frac { \pi }{ 6 } \right) \)
\(\left[ \because { \quad sin }^{ -1 }(sin\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
\(=\pi -\frac { \pi }{ 3 } +\frac { \pi }{ 3 } \quad \left[ \because \quad { cos }^{ -1 }(cos\quad \theta )=\theta \forall \theta (0,\quad \pi ) \right] \)
\(=\pi\)

\(cot\left( \frac { \pi }{ 2 } -2cot^{ -1 }\sqrt { 3 } \right) =\sqrt { 3 } \)
Alternative Method :
\(cot\left( \frac { \pi }{ 2 } -2{ cot }^{ -1 }\sqrt { 3 } \right) =cot\left[ \frac { \pi }{ 2 } -2{ cot }^{ -1 }\left( cot\frac { \pi }{ 6 } \right) \right] \)
\(=cot\left[ \frac { \pi }{ 2 } -2\left( \frac { \pi }{ 6 } \right) \right] \)
\(\left[ \because { cot }^{ -1 }(cot\quad \theta )=\theta \ \forall \theta \ \in \ (0,\quad \pi ) \right] \)
\(=cot\left[ \frac { \pi }{ 2 } -\frac { \pi }{ 3 } \right] \)
\(=cot\left( \frac { \pi }{ 6 } \right) \)
\(=\sqrt { 3 } \)