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

Let \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) =y\), Then \(\sin y=\frac{1}{\sqrt{2}}\)
We know that the range of the principal value branch of \(\sin ^{-1} \text { is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\)
Therefore, principal value of \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) is \ \frac { \pi }{ 4 } \)

Given expression \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Putting \(\sin ^{-1}\left(\frac{3}{5}\right)=x \text { and } \cot ^{-1}\left(\frac{3}{2}\right)=y\)
Or \( \sin (x)=3 / 5 \ and \ \cot y=3 / 2 \)
\(Now, \sin (x)=3 / 5 \Rightarrow \cos x=\sqrt{1-\sin ^2 x}=4 / 5\ and \ \sec x=5 / 4 \)
(Using identities \(\cos \mathrm{x}=\sqrt{1-\sin ^2 x} \text { and } \sec \mathrm{x}=1 / \cos\))
\(\tan x=\sqrt{\sec ^2 x-1}=\sqrt{\frac{25}{16}-1}=3 / 4 \text { and } \tan y=1 / \cot (y)=2 / 3\)
we can written as
\( \tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\cot ^{-1} \frac{3}{2}\right)=\tan (x+y) \)
\(=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \)
\(=17 / 6\)

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

\({ tan }^{ -1 }\left[ \frac { \sqrt { 1+sin\quad x } +{ \sqrt { 1-sin\quad x } }\quad }{ \sqrt { 1+sin\quad x } +{ \sqrt { 1-sin\quad x } } } \right] \)
\(\begin{cases} \because 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 },0\le x\le \frac { \pi }{ 2 } \\ and\quad 1-sin\quad x={ \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } +\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } }{ \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } -\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } +sin\frac { x }{ 2 } +cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } -cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left( \frac { 2cos\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( cot\frac { x }{ 2 } \right) \)
\(={ tan }^{ -1 }tan\left( \frac { \pi }{ 2 } -\frac { x }{ 2 } \right) =\left( \frac { \pi }{ 2 } -\frac { x }{ 2 } \right) \)

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

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
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.