Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

L.H.S = \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 1 }{ 7 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 7 } \right) \left( \frac { 1 }{ 13 } \right) } \right) ={ tan }^{ -1 }\left( \cfrac { \frac { 13+7 }{ 91 } }{ \frac { 91-1 }{ 91 } } \right) \)

\(=\tan ^{-1}\left(\frac{\frac{20}{91}}{\frac {90}{91}}\right)=\tan ^{-1}\left(\frac{20}{\not 91} \times \frac{\not 91}{90}\right) \)
\(=\tan ^{-1}\left(\frac{\not 20^{2}}{\not 90^{9}}\right)=\tan ^{-1}\left(\frac{2}{9}\right)=\mathrm{RHS} \)
Hence proved.

\(cos\left( { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
\({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow \frac { 4 }{ 5 } =cos\theta \)
Also \({ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ sin }^{ -1 }\left( cos\theta \right) \) [using (1)]
= \({ sim }^{ -1 }\left( sin\left( \frac { \pi }{ 2 } -\theta \right) \right) \) \(\left[ \because cos\theta =sin\left( \frac { \pi }{ 2 } -\theta \right) \right] \)
= \(\frac { \pi }{ 2 } -\theta \)
\(\therefore cos\left( cos^{ i1 }\left( \frac { 4 }{ 5 } \right) +sin^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
= \(cos\left( \theta +\frac { \pi }{ 2 } -\theta \right) \)[using (1) & (2)]
= \(cos\frac { \pi }{ 2 } \)
= 0

Let y = sin^-1x
When y = 0, 0 = sin^-1Ix
\(\Rightarrow\) sin(0) = sin (sin^-1)(x))
\(\Rightarrow\)sin 0 = x
\(\Rightarrow\)x = 0
Hence, solution to (1) is x = 0. Also, graph of sin x and sin^-1x intersect at origin (0, 0).

cos (tan^-1x) = \(sin\left( { cot }^{ -1 }\frac { 3 }{ 4 } \right) \)
\(\Rightarrow sin\left( { tan }^{ -1 }\frac { 4 }{ 3 } \right) =sin\left( { sin }^{ -1 }\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right) \)
\(\left[ \because { tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right] \)
\(\Rightarrow sin\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)
\(\Rightarrow cos\left( { tan }^{ -1 }x \right) =cos\left(- { tan }^{ -1 }x \right) =\frac { 4 }{ 5 } \)
\(\left[ \because cosx=cos(-x) \right] \)
\(\Rightarrow { tan }^{ -1 }x={- tan }^{ -1 }x={ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
\(\Rightarrow \tan ^{-1} x=-\tan ^{-1} x=\tan ^{-1} \frac{3}{4}\)
\(\left[ \because { cos }^{ -1 }x={ tan }^{ -1 }\sqrt { \frac { 1-{ x }^{ 2 } }{ x } } ;{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ tan }^{ -1 }\sqrt { \frac { 1-\frac { 16 }{ 25 } }{ \frac { 4 }{ 5 } } } \right] \)
\(\Rightarrow x=\frac { 3 }{ 4 } ,\frac { 3 }{ 4 } \) 

\({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) +{ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 3 } \)
\(\left[ \because { co }^{ -1 }\left( \frac { 1 }{ x } \right) ={ tan }^{ -1 }\left( x \right) \right] \)
\(\Rightarrow 2{ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 3 } \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 6 } \)
\(\Rightarrow \cfrac { 2x }{ 1-{ x }^{ 2 } } =tan\left( \cfrac { \pi }{ 6 } \right) =\cfrac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow 2\sqrt { 3x } =1-{ x }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+2\sqrt { 3x } -1=0\)
\(\Rightarrow x=\frac { -2\sqrt { 3 } \pm \sqrt { 12-4(1)(-1) } }{ 2 } \) \(\left[\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right]\)
\(\Rightarrow x=\cfrac { -2\sqrt { 3 } \pm \sqrt { 16 } }{ 2 } \Rightarrow x=\cfrac { -2\sqrt { 3 } \pm 4 }{ 2 } \)
\(\Rightarrow x=2\left( \frac { -\sqrt { 3 } \pm 2 }{ 2 } \right) \)
\(\Rightarrow x=-\sqrt { 3 } \pm 2\)
\(\Rightarrow x=-2-\sqrt { 3 } \text { or} -2-\sqrt { 3 } \)
Since \(x>0,x=-2-\sqrt { 3 } \) is not possible
\(\therefore x=2-\sqrt { 3 } \)

Let cot−1\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) =\alpha \). Then, cot \(\alpha =\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \) and α is acute.
We construct a right triangle with the given data.
From the triangle, sec\(\alpha=\frac{x}{1}=x\). Thus, \(\alpha\) = sec^-1x
Hence, cot⁻¹\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

Given \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)
\((cos2\frac { \pi }{ 2 } <3x-1\))
\(\Rightarrow 0<3x-1<-1\)
\(0+1<3x<-1+1\)
\(1<3x<0\)
This inequality is true, only when 0 < x < \(\frac { 1 }{ 3 }\)

\(={ sin }^{ -1 }\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \)
= \({ sin }^{ - }\left( sin\left( \frac { 5\pi }{ 9 } +\frac { \pi }{ 9 } \right) \right) \)
(\(\because \) sin A cos B + cos A sin B = sin (A + B))
= \({ sin }^{ -1 }\left( sin\left( \frac { 6\pi }{ 9 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) \) \(\left[ \because \frac { 2\pi }{ 3 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
= \({ sin }^{ -1 }\left( sin\frac { \pi }{ 3 } \right) \) \(\left( \because sin\left( \pi -\theta \right) =sin\theta \right) \)
= \(\frac { \pi }{ 3 } \) \(\left[ \because \frac { \pi }{ 3 } \quad \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \) 

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

\({ tan }^{ -1 }\left( { \frac { x-1 }{ x-2 } } \right) +{ tan }^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }\left( \cfrac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\left( \frac { x-1 }{ x-2 } \right) \left( \frac { x+1 }{ x+2 } \right) } \right) =\frac { \pi }{ 4 } \)

\(\Rightarrow \frac { 2{ x }^{ 2 }-4 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } =1\)
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow \) 2x²- 4 = -3
\(\Rightarrow\) 2x² = -3 + 4 = 1
\(\Rightarrow\) \({ x }^{ 2 }=\frac { 1 }{ 2 } \)
\(\Rightarrow\) \(x=\frac { 1 }{ \sqrt { 2 } } \)

We know that \(sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) =cos^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
Thus, \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =\frac { 1 }{ \sqrt { 1+x^{ 2 } } } \) ...(1)
Let \(\cot ^{-1}\left(\frac{3}{4}\right)=\theta\). Then \(\cot \theta=\frac{3}{4}\) and so \(\theta\) is cute. 
From the diagram, we get,
Hence \(\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}=\sin \theta=\frac{4}{5}\) ................ (2)
Using (1) and (2) in the given equation, we \(\frac { 1 }{ \sqrt { 1+x^{ 2 } } } =\frac { 4 }{ 5 } \) \(\sqrt{1+x^2}=\frac{5}{4}\)
Thus, x = \(\pm\frac{3}{4}\)

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)