\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }x\)             [Given]
Put, x = tan \(\theta \)        \(\Rightarrow\) tan^-1 x=\(\theta \)
\(\Rightarrow \quad { tan }^{ -1 }\left[ \frac { 1-tan\theta }{ 1+tan\theta } \right] =\frac { 1 }{ 2 } \theta \)
\(\Rightarrow \quad { tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\theta \right) \right] =\frac { \theta }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 4 } -\theta =\frac { \theta }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 4 } =\frac { \theta }{ 2 } +\theta \)
\(\Rightarrow \quad \frac { \pi }{ 4 } =\frac { 3\theta }{ 2 } \Rightarrow \theta =\frac { \pi }{ 6 } \)
\(\Rightarrow \quad { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow \quad x=tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)         

Let x = cos 2\(\theta \)
\(=sin\left[ 2t{ an }^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \right] \)
\(=sin\left[ 2tan^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \right] \)
\(\left[ \because cos2\theta =1-2{ sin }^{ 2 }\theta \ and\ cos\ 2\theta =2{ cos }^{ 2 }\theta -1 \right] \)
\(=sin\left[ 2{ tan }^{ -1 }\left( tan\quad \theta \right) \right] \)
\(=sin(2\theta )=\sqrt { 1-{ cos }^{ 2 }2\theta } \)
\(=sin\quad 2\theta =\sqrt { 1-{ x }^{ 2 } } \)

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

\(L.H.S.={ sin }^{ -1 }\frac { 5 }{ 13 } +{ cos }^{ -1 }\frac { 3 }{ 5 } \)
\(\because { sin }^{ -1 }x={ tan }^{ -1 }\left( \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
\({ sin }^{ -1 }\frac { 5 }{ 13 } ={ tan }^{ -1 }\left( \frac { 5/13 }{ \sqrt { 1-25/169 } } \right) \)
\(={ tan }^{ -1 }\frac { (5/13) }{ \sqrt { \frac { 144 }{ 169 } } } ={ tan }^{ -1 }\frac { 5 }{ 12 } \)
\(and\quad { cos }^{ -1 }x={ tan }^{ -1 }\left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ x } \right) \)
\(\therefore \quad { cos }^{ -1 }\frac { 3 }{ 5 } ={ tan }^{ -1 }\left( \frac { \sqrt { 1-\frac { 9 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \sqrt { \frac { 16 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
\({ tan }^{ -1 }\frac { 5 }{ 12 } +{ tan }^{ -1 }\frac { 4 }{ 3 } ={ tan }^{ -1 }\left[ \frac { \frac { 5 }{ 12 } +\frac { 4 }{ 2 } }{ 1-\frac { 5\times 4 }{ 12\times 3 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left| \frac { \frac { 15+48 }{ 36 } }{ \frac { 36-20 }{ 36 } } \right| \)
\(={ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =R.H.S.\)

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

\(sin\left( 2sin^{ -1 }\frac { 3 }{ 5 } \right) =\frac { 24 }{ 25 } \)
Alternative Method :
\(sin\left( 2sin^{ -1 }\frac { 3 }{ 5 } \right) =sin\left( 2tan^{ -1 }\frac { 3 }{ 4 } \right) \)
\(=sin\left[ { tan }^{ -1 }\left( \frac { 2\times \frac { 3 }{ 4 } }{ 1-\frac { 9 }{ 16 } } \right) \right] \)
\(=sin\left[ { tan }^{ -1 }\left( \frac { 3 }{ 2 } \times \frac { 16 }{ 7 } \right) \right] \)
\(=sin\left[ { tan }^{ -1 }\frac { 24 }{ 7 } \right] \)
\(=sin\left[ { sin }^{ -1 }\frac { 24 }{ 25 } \right] \)
\(=\frac { 24 }{ 25 } \)
\(\\ \left[ \because { sin }^{ -1 }(sin\quad \theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)

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

\( y=\operatorname{cosec}^{-1}(-\sqrt{2}) \)
\(\Rightarrow \operatorname{cosec} y=-\sqrt{2} \)
\(\Rightarrow \operatorname{cosec} y=\operatorname{cosec}\left(-\frac{\pi}{4}\right) \)
We know that the range of the principal value branch of cosec⁻¹x is \( {\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} .} \)
\(\therefore y=-\frac{\pi}{4} \)
Hence, the principal value of  \(\operatorname{cosec}^{-1}(-\sqrt{2}) \text { is }-\frac{\pi}{4}\)

\({ R }_{ 1 }\rightarrow \frac { 1 }{ a } { R }_{ 1 },{ R }_{ 2 }\rightarrow \frac { 1 }{ b } ,{ R }_{ 3 }\rightarrow \frac { 1 }{ c } { R }_{ 3 }\)
\(LHS=abc\left| \begin{matrix} \frac { 1 }{ a } +1 & \frac { 1 }{ a } & \frac { 1 }{ a } \\ \frac { 1 }{ b } & \frac { 1 }{ b } +1 & \frac { 1 }{ b } \\ \frac { 1 }{ c } & \frac { 1 }{ c } & \frac { 1 }{ c } +1 \end{matrix} \right| \)
\(=abc\left| \begin{matrix} 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } & 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } & 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \\ \frac { 1 }{ b } & \frac { 1 }{ b } +1 & \frac { 1 }{ b } \\ \frac { 1 }{ c } & \frac { 1 }{ c } & \frac { 1 }{ c } +1 \end{matrix} \right| \)
Taking \(\left( 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \right) \) common form R₁
\(=abc\left( 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \right) \left| \begin{matrix} 1 & 1 & 1 \\ \frac { 1 }{ b } & \frac { 1 }{ b } +1 & \frac { 1 }{ b } \\ \frac { 1 }{ c } & \frac { 1 }{ c } & \frac { 1 }{ c } +1 \end{matrix} \right| \)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }-C_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-C_{ 1 }\)
\(=abc\left( 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \right) \left| \begin{matrix} 1 & 0 & 0 \\ \frac { 1 }{ b } & 1 & 0 \\ \frac { 1 }{ c } & 0 & 1 \end{matrix} \right| \)
Expanding C₂
\(=abc\left( 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \right) (1)\)
\(=abc+bc+ca+ab=RHS\)

Operating C₁ \(\rightarrow\) C₁-(C₂+C₃)
LHS=\(\left| \begin{matrix} -2a & c+a & a+b \\ -2p & r+p & p+q \\ -2x & z+x & x+y \end{matrix} \right| \)
Taking (-2) Common from
C₁=-2\(\left| \begin{matrix} a & c+a & a+b \\ p & r+p & p+q \\ x & z+x & x+y \end{matrix} \right| \)
C₂ \(\Rightarrow\) C₂-C₁, C₃ \(\Rightarrow\) C₃-C₁
LHS=-2\(\left| \begin{matrix} a & c & b \\ p & r & q \\ x & z & y \end{matrix} \right| \)
\({ C }_{ 2 }\leftrightarrow { C }_{ 3 }=+2\left| \begin{matrix} a & b & c \\ p & q & r \\ x & y & z \end{matrix} \right| =RHS\)

After dividing numerator and denominator by a^3 we have,
\( \tan ^{-1}\left(\frac{3\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2}\right) \)
\(Put \mathrm{x} / \mathrm{a}=\tan \theta\ and\ \theta=\tan ^{-1}(\mathrm{x} / \mathrm{a}) \)
\( =\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right) \)
\(=\tan ^{-1}(\tan 3 \theta) \)
\(=3 \theta \\ =3 \tan ^{-1}(\mathrm{x} / \mathrm{a})\)

We know that sin⁻¹ (sin x) = x . Therefore, \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\frac{3 \pi}{5}\)
but \(\frac{3 \pi}{5} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) which is the principal branch of sin^–1 x
However \(\sin \left(\frac{3 \pi}{5}\right)=\sin \left(\pi-\frac{3 \pi}{5}\right)=\sin \frac{2 \pi}{5} \text { and } \frac{2 \pi}{5} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) 
Therefore \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)=\frac{2 \pi}{5}\)

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 \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =\frac { 2\pi }{ 3 } .\)