LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)

\({ sin }^{ -1 }\left( \frac { sinx+cosx }{ \sqrt { 2 } } \right)\)
\(\begin{aligned} & =\sin ^{-1}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right) \\ & =\sin ^{-1}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right) \\ & =\sin ^{-1}\left(\sin \left(x+\frac{\pi}{4}\right)\right)=\pi+\frac{\pi}{4} \end{aligned}\)

Put \(x=a\ cos\theta\)
\(f(x)={ tan }^{ -1 }\sqrt { \frac { a-acos\theta }{ a+acos\theta } } ={ tan }^{ -1 }\sqrt { \frac { 1-cos\theta }{ 1+cos\theta } }\)
= \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\frac { \theta }{ 2 } }{ 2{ cos }^{ 2 }\frac { \theta }{ 2 } } } =tan|tan\frac { \theta }{ 2 } |={ tan }^{ -1 }\left( tan\frac { \theta }{ 2 } \right)\)
= \(\frac { \theta }{ 2 }\) \(\left.\Rightarrow 0<\frac{\theta}{2}<\frac{\pi}{2} \therefore\left|\tan \frac{\theta}{2}\right|=\tan \frac{\theta}{2}\right]\)
\(\frac { 1 }{ 2 } .{ cos }^{ -1 }\left( \frac { x }{ a } \right)\)

The domain of sin^-1x is [-1, 1]
\(\therefore\) f(x) = sin^-1(2x - 3) is defined for all x, satisfying
\(-1\le 2x-3\le 1\)
\(\Rightarrow 3-1\le 2x\le 1+3\)
\(\Rightarrow 2\le 2x\le 4\Rightarrow 1\le x\le 2\Rightarrow x\epsilon \left[ 1,2 \right] \)
\(\therefore\) Domain of f(x) = sin^-1(2x - 3) is [1, 2].
(ii) The domain of f(x) is [-1, 1] and that of cosx is R
\(\therefore\) Domain of f(x) = sin^-1x + cos x is
\(\left[ -1,1 \right] \cap R=\left[ -1,1 \right] \)