\(\sin \left(\frac{\pi}{3}+\cos ^{-1}\left(\frac{-1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{2 \pi}{3}\right)\)
= sin \(\pi\)
= 0

We have \(\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right) \neq \frac{13 \pi}{6}\) it does not lie between 0 and \(\pi\)
\( \cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right) =\cos ^{-1}\left(\cos \left(2 \pi+\frac{\pi}{6}\right)\right) \)
\(=\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\frac{\pi}{6} \)
\(\therefore\) The principle value of \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \text { is } \frac{\pi}{6}\)

Let \(\operatorname{cosec}^{-1}(-\sqrt{2})\) = y, where \(y \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\{-0\} \)
\( \Rightarrow \operatorname{cosec} y= -\sqrt{2} \)
\(\Rightarrow \operatorname{cosec} y= \operatorname{cosec}\left(\frac{-\pi}{4}\right) \) \([\because \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta]\)
\( y =\frac{-\pi}{4} \)
The principal value of \(\operatorname{cosec}^{-1}(-\sqrt{2}) \text { is }-\left(\frac{\pi}{4}\right)\)
\(\Rightarrow y=\frac{-\pi}{4}\)

Let \( \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=y , \) where \(\frac{-\pi}{2}<y<\frac{\pi}{2} \)
\(\Rightarrow \sin y=\frac{1}{\sqrt{2}} \Rightarrow \sin y=\sin \frac{\pi}{4} \Rightarrow y=\frac{\pi}{4} \)
The principal value of \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)

Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =\theta \Rightarrow cos\theta =\frac { 1 }{ 2 } \)
\(\Rightarrow sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 4 } } =\sqrt { \frac { 3 }{ 4 } } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) =\frac { \sqrt { 3 } }{ 2 } \)

LHS = \(2{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } +\cfrac { 2 }{ 3 } }{ 1-\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } \right) } \right) ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } }{ \frac { 9-2 }{ 9 } } \right) \)
= \({ tan }^{ -1 }\left( \frac { 4 }{ 3 } \times \frac { 9 }{ 7 } \right) ={ tan }^{ -1 }\left( \frac { 12 }{ 7 } \right) \)
= RHS
Hence proved

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow cos\theta =\frac { 4 }{ 5 } \)
\(\therefore sin\left( \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) =sin\left( \frac { 1 }{ 2 } \theta \right) \)
= \(sin\frac { \theta }{ 2 } =\sqrt { \frac { 1-cos\theta }{ 2 } } \)
= \(\sqrt { \frac { 1-\frac { 4 }{ 5 } }{ 2 } } =\sqrt { \frac { 5-4 }{ 5(2) } } =\sqrt { \frac { 1 }{ 10 } } \)

Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

L.H.S = \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 1 }{ 7 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 7 } \right) \left( \frac { 1 }{ 13 } \right) } \right) ={ tan }^{ -1 }\left( \cfrac { \frac { 13+7 }{ 91 } }{ \frac { 91-1 }{ 91 } } \right) \)

\(=\tan ^{-1}\left(\frac{\frac{20}{91}}{\frac {90}{91}}\right)=\tan ^{-1}\left(\frac{20}{\not 91} \times \frac{\not 91}{90}\right) \)
\(=\tan ^{-1}\left(\frac{\not 20^{2}}{\not 90^{9}}\right)=\tan ^{-1}\left(\frac{2}{9}\right)=\mathrm{RHS} \)
Hence proved.

Given 
\({ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) ={ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore x=\frac { 1 }{ \sqrt { 3 } } \)

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\) where \(0\le y\le \pi \)
Then \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\Rightarrow cosy=\frac { -1 }{ 2 } \)
\(\Rightarrow cos\ y=-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\left( \frac { 2\pi }{ 3 } \right) \)
\(\Rightarrow y=\frac { 2\pi }{ 3 } \left[ \because \frac { 2\pi }{ 3 } \in \left[ 0,\pi \right] \right] \)
\(\therefore \) The principal value of \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) \frac { 2\pi }{ 3 } \)

Let sin-1(-1) = y where \(\frac { -\pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
Then \(sin^{ -1 }(-1)=y\Rightarrow sin\quad y=-1\)
\(-1=sin\left( \frac { -\pi }{ 2 } \right) \Rightarrow y=\frac { -\pi }{ 2 } \) \(\left[ \because \frac { -\pi }{ 2 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\( \therefore\)  The principal value of \({ sin }^{ -1 }(-1)\ is \ \frac { -\pi }{ 2 } \)

Let \({ tan }^{ -1 }\left( \frac { -1 }{ \sqrt { 3 } } \right) =y\) where \(\frac{-\pi}{2}<y<\frac{\pi}{2}\)
\(\Rightarrow tan\ y=\frac { -1 }{ 3 } =-tan\frac { \pi }{ 6 } =tan\left( \frac { -\pi }{ 6 } \right) \)
\(y=\frac { -\pi }{ 6 } \) \(\left[ \because \frac { -\pi }{ 6 } \epsilon \left( \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
The principal value of \({ tan }^{ -1 }\left( \frac { -1 }{ \sqrt { 3 } } \right) =6\)

sin x = -1
\(\Rightarrow sinx=sin\left( \frac { -\pi }{ 2 } \right) \Rightarrow x=\frac { -\pi }{ 2 } ,\frac { 3\pi }{ 2 } ,\frac { 7\pi }{ 2 } ...\)
\(\Rightarrow x=(4n-1)\frac { \pi }{ 2 } ,n\varepsilon Z.\)
\(\Rightarrow x=(4n-1)\frac { \pi }{ 2 } \)
n takes the values \(0,\pm 1,\pm 2,\pm 3and\pm 4\)
since when n = -4, \(x=\frac { -17\pi }{ 2 } <-8\pi \)