\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right) \)
\(\text { Put } x^{2}=\cos 2 \theta, \text { then } \)
\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)\) 
\( \tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right)\) 
\(\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)\)
[divide numerator and denominator inside the bracket by \(cos \theta]\)

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
if \(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) <1\)
\(\Rightarrow tan^{ -1 }\left[ \frac { x(x+1)+(x-1)^{ 2 } }{ \left( x-1 \right) x-\left( x+1 \right) \left( x-1 \right) } \right] =tan^{ -1 }7\)
\(\Rightarrow \frac { \left( x^{ 2 }+x \right) +\left( x^{ 2 }+1-2x \right) }{ \left( x^{ 2 }-x \right) -\left( x^{ 2 }-1 \right) } =tan\left[ -tan^{ -1 }7 \right] \)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \) 2x² - 8x + 8 = 0
\(\Rightarrow \) (x - 2)² = 0
\(\Rightarrow \) x = 2
Let us now verify whether x = 2 satisfies the condition (i)
\(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) =3\times \frac { 1 }{ 2 } =\frac { 3 }{ 2 } \) Which is not less than 1.
Hence this value does not satisfy the condition (i) there is no solution to the given trigonometric equation.

\( \text { We have, } \tan \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right) \)
\(=\tan \frac{1}{2}\left(2 \tan ^{-1} x+2 \tan ^{-1} y\right) \) 
 \(\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right] \)
\(= \tan \frac{1}{2} \times 2\left(\tan ^{-1} x+\tan ^{-1} y\right) \)
\(= \tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \)
\(= \frac{x+y}{1-x y} \because \tan ^{-1} x+\tan ^{-1} y=\mid \tan ^{-1}\left(\frac{x+y}{1-x y}\right) \)
\(\text { and } \tan \left(\tan ^{-1} \theta\right)=\theta \)

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 \({ cot }^{ -1 }\left( \sqrt { 3 } \right) =y,0
\(\Rightarrow coty=\sqrt { 3 } \Rightarrow y=\frac { \pi }{ 6 } \)
Hence, the required principal value  \(\frac { \pi }{ 6 } \)

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 = \(-\frac { \pi }{ 3 } \)

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

\(LHS={ tan }^{ -1 }\frac { 1 }{ 2 } +{ tan }^{ -1 }\frac { 2 }{ 11 } \)
\(={ tan }^{ -1 }\frac { \frac { 1 }{ 2 } +\frac { 2 }{ 11 } }{ 1-\frac { 1 }{ 2 } .\frac { 2 }{ 11 } } ={ tan }^{ -1 }\frac { 15 }{ 20 } ={ tan }^{ -1 }\frac { 3 }{ 4 } =RHS.\)

\({ sin }^{ -1 }\left[ \frac { x+\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 2 } } \right] \quad Let\quad x=sin\theta \Rightarrow \theta ={ sin }^{ -1 }x\)
\(={ sin }^{ -1 }\left( \frac { sin\quad \theta +\sqrt { 1-{ sin }^{ 2 } } \theta }{ \sqrt { 2 } } \right) \)
\(={ sin }^{ -1 }\left( \frac { sin\theta +cos\theta }{ \sqrt { 2 } } \right) \)
\(={ sin }^{ -1 }\left( sin\theta \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } cos\theta \right) \)
\(={ sin }^{ -1 }\left( sin\theta cos\frac { \pi }{ 4 } +cos\theta sin\frac { \pi }{ 4 } \right) \)
\(={ sin }^{ -1 }\left[ \theta +\frac { \pi }{ 4 } \right] \)
\(\Rightarrow \theta +\frac { \pi }{ 4 } =\frac { \pi }{ 4 } +{ sin }^{ -1 }x\)

\({ sin }^{ -1 }x=A\ and\ { sin }^{ -1 }y=B\)
\(\Rightarrow\) x = sin A and y = sin B
\(\therefore cos\ A=\sqrt { 1-{ x }^{ 2 } } ,cos\ B=\sqrt { 1-{ y }^{ 2 } } \)
we have sin (A+B) = sin A cos B + cos A sin B
\(\Rightarrow sin(A+B)=x\sqrt { 1-{ y }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } y\)
\(\Rightarrow sin(A+B)=x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \)
\(\Rightarrow A+B={ sin }^{ -1 }\left( x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \right) \)
\(\therefore { sin }^{ -1 }x+sin^{ -1 }y={ sin }^{ -1 }\left( x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \right) \)

\(L.H.S.={ tan }^{ -1 }\frac { 2 }{ 3 } =\frac { 1 }{ 2 } \left( 2{ tan }^{ -1 }\frac { 2 }{ 3 } \right) \)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }\left( \frac { 2\times \frac { 2 }{ 3 } }{ 1-\frac { 4 }{ 9 } } \right) \right] \)
\(\left[ \because 2{ tan }^{ -1 }x={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left[ \frac { \frac { 4 }{ 3 } }{ \frac { 9-4 }{ 9 } } \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 4\times 9 }{ 3\times 5 } \right) \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 12 }{ 5 } \right) =R.H.S.\)

\({ tan }^{ -1 }\frac { x }{ y } -{ tan }^{ -1 }\frac { x-y }{ x+y } \)
\(={ tan }^{ -1 }\left[ \frac { \frac { x }{ y } -\frac { x-y }{ x+y } }{ 1+\left( \frac { x }{ y } \right) \left( \frac { x-y }{ x+y } \right) } \right] \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\left[ \frac { \frac { x(x+y)-y(x-y) }{ y(x+y) } }{ \frac { y(x+y)-x(x-y) }{ y(x+y) } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { { x }^{ 2 }+xy-xy+{ y }^{ 2 } }{ { xy+y }^{ 2 }+{ x }^{ 2 }-xy } \right] \)
\(={ tan }^{ -1 }\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) ={ tan }^{ -1 }1=\frac { \pi }{ 4 } \)   Hence Proved

\({ tan }^{ -1 }\left[ \frac { cos\quad x }{ 1+sin\quad s } \right] \) \(\quad \because \) \(\begin{cases} cos\quad x={ cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } \\ and\quad 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { { cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] Divide\quad by\quad cos\frac { x }{ 2 } ,\quad we\quad get\)
\(={ tan }^{ -1 }\left[ \frac { 1-tan\frac { x }{ 2 } }{ 1+tan\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) \right] =\frac { \pi }{ 4 } -\frac { x }{ 2 } \)