Firstly, assume the given inverse trigonometric function equal to y( or x) Let \(y=\cos ^{-1}\left(-\frac{1}{2}\right)\)
\(\Rightarrow \cos y=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right) \)
\(\Rightarrow \cos y=\cos \left(\frac{2 \pi}{3}\right) \Rightarrow y=\frac{2 \pi}{3}\)
Since, the principal value branch of \(\cos ^{-1} is [0, \pi]\) and
\(\frac{2 \pi}{3} \in[0, \pi]\) 
So, the principal value of \(\cos ^{-1}\left(-\frac{1}{2}\right)\) is \(\frac{2 \pi}{3}\).
Now, find the principal value of \(tan ^{-1}(-\sqrt{3})\ Let\ x=\tan ^{-1}(-\sqrt{3})\)
\(\Rightarrow \tan x=-\sqrt{3}=-\tan \frac{\pi}{3} \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]\)
\(\Rightarrow \tan x=\tan \left(\frac{-\pi}{3}\right) \quad[\because \tan (-\theta)=-\tan \theta]\)
\(\Rightarrow x=-\frac{\pi}{3}\)
Since, the principal value branch of \(\tan ^{-1}\) is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(-\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) 
Now, find the principal value of \(\operatorname{cosec}^{-1}\) 
Let \(z=\operatorname{cosec}^{-1}(2) \Rightarrow \operatorname{cosec} z=2=\operatorname{cosec} \frac{\pi}{6} \Rightarrow z=\frac{\pi}{6}\)
Since, the principal value branch of \(\operatorname{cosec}^{-1}\) is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\) and \(\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)
So, the principal value of \(\operatorname{cosec}^{-1}(2)\) is \(\frac{\pi}{6}\).
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)+\tan ^{-1}(-\sqrt{3})-\operatorname{cosec}^{-1} 2\)
\(=\frac{2 \pi}{3}-\frac{\pi}{3}-\frac{\pi}{6}=\frac{(4-2-1) \pi}{6}=\frac{\pi}{6}\)

\(cos\left[ { tan }^{ -1 }(x) \right] =sin\left[ { cot }^{ -1 }\frac { 3 }{ 4 } \right] \)
\(\Rightarrow \quad cos\left[ { cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right] =sin\left[ { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right] \)
\(\Rightarrow \quad \quad \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } =\frac { 4 }{ 5 } \)
\(\Rightarrow \quad \quad 1+{ x }^{ 2 }=\frac { 25 }{ 16 } \)
\(\Rightarrow \quad \quad { x }^{ 2 }=\frac { 25 }{ 16 } -1=\frac { 9 }{ 16 } \)
\(\Rightarrow \quad \quad x=\frac { 3 }{ 4 } ,-\frac { 3 }{ 4 } \)
\(\Rightarrow \quad x=-\frac { 3 }{ 4 } \) does not satisfy so \(x=\frac { 3 }{ 4 } \)

Let, \({ cos }^{ -1 }x=\alpha \Rightarrow x=cos\alpha \)
LHS=\(\alpha +{ cos }^{ -1 }\left[ cos\alpha cos\left( \frac { \pi }{ 3 } \right) +\frac { \sqrt { 3 } }{ 2 } \sqrt { 1-{ cos }^{ 2 }\alpha } \right] \)
\(=\alpha +{ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 3 } \right) cos\alpha +sin\frac { \pi }{ 3 } sin\alpha \right] \)
\(=\alpha +{ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 3 } -\alpha \right) \right] \)
\(=\alpha +\frac { \pi }{ 3 } -\alpha \)
\(\left[ \because \quad { cos }^{ -1 }(cos\theta )=\theta \forall \theta \in (0,\pi ) \right] \)
\(=\frac { \pi }{ 3 } =RHS\)

\({ cot }^{ -1 }\left( \frac { \sqrt { 1+sin\quad x } +\sqrt { 1-sin\quad x } }{ \sqrt { 1+sin\quad x } -\sqrt { 1-sin\quad x } } \right) \)
\(={ cot }^{ -1 }\left[ \frac { \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +{ sin }^{ 2 }\frac { x }{ 2 } +2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) + } }{ \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +{ sin }^{ 2 }\frac { x }{ 2 } +2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) - } } \frac { \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +{ sin }^{ 2 }\frac { x }{ 2 } -2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) } }{ \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +sin^{ 2 }\frac { x }{ 2 } -2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) } } \right] \)
\(={ cot }^{ -1 }\left[ \frac { \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } +\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } }{ \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 }-{ \sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } } } } \right] \)
\(={ cot }^{ -1 }\left[ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) +\left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) -\left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) } \right] \)
\(={ cot }^{ -1 }\left[ \frac { 2cos\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } } \right] \left[ \because \quad { cot }^{ -1 }(cot\quad \theta )=\forall \theta \in (0,\quad N) \right] \)
\(={ cot }^{ -1 }\left[ cot\frac { x }{ 2 } \right] \)
\(=\frac { x }{ 2 } \)

\(\Rightarrow \tan ^{-1}\left(\frac{3 x+2 x}{1-3 x \times 2 x}\right)=\frac{\pi}{4}\)
\(\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right),\right. if \left.x y<1\right]\)
\(\Rightarrow\)\(\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\frac{\pi}{4}\)
\(\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4}\)
\(\left[\because \tan ^{-1}(\theta)=\phi \Rightarrow \theta=\tan \phi\right]\)
\(\Rightarrow \frac{5 x}{1-6 x^{2}}=1\)
\(\Rightarrow\)\(5 x=1-6 x^{2}\)
\(\Rightarrow 6 x^{2}+6 x-x-1=0\) 
\(\Rightarrow 6 x(x+1)-1(x+1)=0\)
\(\Rightarrow(6 x-1)(x+1)=0\)
\( 6 x-1=0 \Rightarrow x=\frac{1}{6}\)
and \(x+1=0 \Rightarrow x=-1\)
Hence, the required value of \(x \text { is } \frac{1}{6}\)

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

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

\(\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)=\sin ^{-1}\left[\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right]\)
\(=\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)=\sin ^{-1}\left[\cos \left(\frac{\pi}{2}+\frac{\pi}{10}\right)\right] \)
\(=\sin ^{-1}\left(-\sin \frac{\pi}{10}\right)=-\sin ^{-1}\left(\sin \frac{\pi}{10}\right)=-\frac{\pi}{10}\)

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

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

\(\cos ^{-1}\left[\cos \left(-\frac{7 \pi}{3}\right)\right]=\cos ^{-1}\left[\cos \frac{7 \pi}{3}\right][\because \cos (-\theta)=\cos \theta] \)
\( =\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{3}\right)\right] \)
\( =\cos ^{-1}\left(\cos \frac{\pi}{3}\right)=\frac{\pi}{3} \in[0, \pi]\)

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 have, \(\sin ^{-1}  \left(\sin \frac{3 \pi}{4}\right)+\cos ^{-1}\left(\cos \frac{3 \pi}{4}\right)+\tan ^{-1}(1) \)
\(=\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{4}\right)\right]+\cos ^{-1}\left[\cos \frac{3 \pi}{4}\right]+\frac{\pi}{4} \)
\(=\sin ^{-1}\left(\sin \frac{\pi}{4}\right)+\frac{3 \pi}{4}+\frac{\pi}{4}\)
\(=\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{\pi}{4}=\frac{5 \pi}{4}\)

We have, \(\tan ^{-1}  {\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]+\tan ^{-1} 1 } \)
\(=\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)\right)\right]+\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \)
\(=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]+\frac{\pi}{4}\)
\(=\tan ^{-1}\left[2 \cos \left(\frac{\pi}{3}\right)\right]+\frac{\pi}{4}\)
\(=\tan ^{-1}\left[2 \times \frac{1}{2}\right]+\frac{\pi}{4}\)
\(=\tan ^{-1}(1)+\frac{\pi}{4}=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}\)

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.

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

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

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

\(\cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right)\) 
\(\Rightarrow \sin \left\{\frac{\pi}{2}-\tan ^{-1} x\right\}=\sin \left(\cot ^{-1} \frac{3}{4}\right)\) 
On equating both sides, we get 
 \( \frac{\pi}{2}-\tan ^{-1} x =\cot ^{-1} \frac{3}{4} \)
\(\Rightarrow \cot ^{-1} x=\cot ^{-1} \frac{3}{4} \quad\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \) 
Again, equating both sides, we get \(x=\frac{3}{4}\)

\( \tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\tan \left[\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right]\) 

\(=\tan \left[\tan ^{-1} \frac{\left(\frac{3}{4}+\frac{2}{3}\right)}{1-\frac{3}{4} \times \frac{2}{3}}\right] \)
\(=\tan \left[\tan ^{-1} \frac{\frac{17}{12}}{\frac{1}{2}}\right] \)
\(=\tan \left[\tan ^{-1} \frac{17}{6}\right]=\frac{17}{6}\)

\( \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)\) 
\(\frac{-\pi}{6}+\frac{\pi}{3}+\tan ^{-1}(-1)\) 
\( {\left[\because \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6}, \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{3}, \sin \left(\frac{-\pi}{2}\right)=-1\right]} \)
\(=\frac{-\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}=\frac{-\pi}{12}\)

\(\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3}) =\tan ^{-1} \sqrt{3}-\left[\pi-\cot ^{-1}(\sqrt{3})\right] \)
\(=\left(\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}\right)-\pi \)
\(=\frac{\pi}{2}-\pi=\frac{-\pi}{2}\)

(i) (b) \(\ln \triangle A B D\)
\(\tan \alpha  =\frac{B D}{A D}\)
\( =\frac{30}{30 \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow  \alpha  =30^{\circ}\)
\(\therefore  \sin \alpha =\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow  \alpha  =\sin ^{-1}\left(\frac{1}{2}\right)\)
(ii) (c) \(\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \alpha =\cos ^{-1}\left(\frac{\sqrt{3}}{2})\right.\)
(iii) (d) \(\ln \triangle B DC\)
\(\tan \beta  =\frac{(BD)}{(C D)}\)
\(=\frac{30}{10 \sqrt{3}}=\sqrt{3} \)
\(\Rightarrow \beta =\tan ^{-1}(\sqrt{3})\)
(iv) \( \text { (c) Since, } \alpha=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \text {, } \)
\(\beta=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)
\(\therefore \quad \angle A B C=\pi-\left(\frac{\pi}{6}+\frac{\pi}{3}\right)\)
\( =\pi-\frac{\pi}{2}=\frac{\pi}{2}\)
(v) (c) We know that domain and range of \(\cos ^{-1} x\) are [-1,1] and \([0, \pi]\) respectively.