Let \({ cosec }^{ -1 }\left( -\sqrt { 2 } \right) =y\), where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] -\left\{ 0 \right\} \)
\(\Rightarrow cosecy=-\sqrt { 2 } =-cosec\frac { \pi }{ 4 } \)
\(=cosec\left( -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

Let's consider \(\tan ^{-1}(1)=x\). Then, \(\tan x=1=\tan \left(\frac{\pi}{4}\right)\). \(\therefore \tan ^{-1}(1)=\frac{\pi}{4}\)
Let's assume,\(\cos ^{-1}\left(-\frac{1}{2}\right)=y\).
Then, \(\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)\)
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)
Let's again assume that \(\sin ^{-1}\left(-\frac{1}{2}\right)=z\).
Then, \(\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)\).
\(\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)
\(\therefore \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)
\(=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \)
\(=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\)

Let tan^-1(-1) = y, where \(y\in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(\Rightarrow tany=-1=-tan\frac { \pi }{ 4 } =tan\left( -\frac { \pi }{ 4 } \right) \)
\( \Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

Let \({ tan }^{ -1 }\left( -\sqrt { 3 } \right) =y\) where \(y\in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(\Rightarrow tany=-\sqrt { 3 } =-tan\left( \frac { \pi }{ 3 } \right) \)
\(=tan\left( -\frac { \pi }{ 3 } \right)\)
\(\Rightarrow \ y=-\frac { \pi }{ 3 } \)
Hence, the reqd. principal value = \(-\frac { \pi }{ 3 } \)

Let \({ cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =y\), where \(y\in [0,\pi ]\)
\(\Rightarrow cosy=\frac { \sqrt { 3 } }{ 2 } \Rightarrow cosy=cos\frac { \pi }{ 6 } \)
\(\Rightarrow \ y=\frac { \pi }{ 6 } \)
Hence, the required Principal value = \(\frac { \pi }{ 6 } \)

We have 
\(\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]=\tan ^{-1}\left[\frac{\frac{a \cos x-b \sin x}{b \cos x}}{\frac{b \cos x+a \sin x}{b \cos x}}\right]=\tan ^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right]\)
\(=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)=\tan ^{-1} \frac{a}{b}-x\)

\(\text {Let } \sin ^{-1} \frac{12}{13}=x, \quad \cos ^{-1} \frac{4}{5}=y, \tan ^{-1} \frac{63}{16}=z\)
\( \sin x=\frac{12}{13}, \quad \cos y=\frac{4}{5}, \quad \tan z=\frac{63}{16}\)
\(\cos x=\frac{5}{13}, \sin y=\frac{3}{5}, \tan x=\frac{12}{5} \text { and } \tan y=\frac{3}{4}\)
\(\text {We have }\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}=-\frac{63}{16}\)
Hence tan(x + y) = − tan z
i.e., tan (x + y) = tan (–z) or tan (x + y) = tan (p – z)
Therefore x + y = – z or x + y = p – z
Since x, y and z are positive, x + y \(\ne\) – z (Why?)
\(\text {Hence }x+y+z=\pi \text { or } \sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi\)

\(\text {Let } \sin ^{-1} \frac{3}{5}=x \text { and } \sin ^{-1} \frac{8}{17}=y\)
\( \sin x=\frac{3}{5} \text { and } \sin y=\frac{8}{17}\)
\(\text {Now }\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
\(\text {and }\cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{64}{289}}=\frac{15}{17}\)
We have cos(x−y) = cos x cos y + sin x siny
\(=\frac{4}{5} \times \frac{15}{17}+\frac{3}{5} \times \frac{8}{17}=\frac{84}{85}\)
\( x-y=\cos ^{-1} \frac{84}{85}\)
\(\text {Hence } \ \sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}\)

We have \(cos({ sec }^{ -1 }x+{ cosec }^{ -1 }x)=cos\left( \frac { \pi }{ 2 } \right) =0\)

Put x = tan \(\theta\) so that \(\theta\) = tan^-1x.
\(RHS={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ={ tan }^{ -1 }\left( \frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \right) \)
\(={ tan }^{ -1 }(tan3\theta )=3\theta =3{ tan }^{ -1 }x\)
\( ={ tan }^{ -1 }x+2{ tan }^{ -1 }x={ tan }^{ -1 }x+{ tan }^{ -1 }\frac { 2x }{ { 1-x }^{ 2 } } =LHS.\)

Let x = sec θ, then \(\sqrt{x^2-1}=\sqrt{\sec ^2 \theta-1}=\tan \theta\)
Therefore, \(\cot ^{-1} \frac{1}{\sqrt{x^2-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x\) which is the simplest form

We write
\( \tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)= \tan ^{-1}\left[\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right] \)
\(= \tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}\right] \)
\(= \tan ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]=\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right] \)
\( =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2} \)

Let sin^-1 x = \(\theta\) then  sin^-1 x = \(\theta\). we have
sin^-1 \((2x\sqrt { 1-{ x }^{ 2 } } )\) = sin^-1\(2\sin { \theta } \sqrt { 1-{ sin }^{ 2 } } \)
= sin^–1 (2sinθ cosθ) = sin^–1 (sin2θ) = 2θ = 2 sin^–1 x

Let  \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =y\), Then \(\cot y=\frac{-1}{\sqrt{3}}=-\cot \left(\frac{\pi}{3}\right)=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \left(\frac{2 \pi}{3}\right)\)
We know that the range of principal value branch of cot^–1 is (0, π) and \(\cot \left(\frac{2 \pi}{3}\right)=\frac{-1}{\sqrt{3}}\)
Hence, principal value of \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =\frac { 2\pi }{ 3 } .\)