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 cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \) = y. Then, cos y = \( \frac { \sqrt { 3 } }{ 2 } \)
The range of the principal values of y = cos⁻¹x is [0, \(\pi\)].
So, let us find y in [0, \(\pi\)] such that cos y =\( \frac { \sqrt { 3 } }{ 2 } \)
But, cos\(\frac{\pi}{6}=\frac{\sqrt3}{2} and \frac{\pi}{6}\in[0,\pi]\). Therefore, y = \(\frac{\pi}{6}\)
Thus, the principal value of cos^-1 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

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).

Given  \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1=sin\frac { \pi }{ 2 } \)
\(\left[ \because sin\frac { \pi }{ 2 } =1 \right] \)
\(\Rightarrow { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x={ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 2 } \right) \right) \)
\(\Rightarrow { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x=\frac { \pi }{ 2 } \)
\({ sin }^{ -1 }\frac { 1 }{ 5 } =\frac { \pi }{ 2 } -{ cos }^{ -1 }x{ sin }^{ -1 }x\)
\(\left[ \because { sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } \right] \)
\(\Rightarrow x=\frac { 1 }{ 5 } \)

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

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 < \(\frac { 1 }{ 3 }\)

cos x = 0
\(\Rightarrow x=\left( 2n+1 \right) \frac { \pi }{ 2 } ,n\varepsilon Z\)
But \(-6\pi \le x\le 6\pi \)
\(\therefore \) n can take values from  
\(x=(2n+1)\frac { \pi }{ 2 } ,n=0\pm 1,\pm 2,...\pm 5\), -6

\(={ sin }^{ -1 }\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \)
= \({ sin }^{ - }\left( sin\left( \frac { 5\pi }{ 9 } +\frac { \pi }{ 9 } \right) \right) \)
(\(\because \) 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

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

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 = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

If x = 0 , then both sides are equal to 0  .... (1)
Assume that 0< x <1.
Let \(\theta\) = sin⁻¹ x. Then 0 < \(\theta<\frac{\pi}{2}\). Now, sin \(\theta=\frac{x}{1}\) gives tan \(\theta=\frac{x}{\sqrt{1-x^2}}\)
Hence, tan (sin^-1 x) = \(\frac{x}{\sqrt{1-x^2}}\) ....... (2)
Assume that −1 < x < 0. Then, \(\theta\) = sin⁻¹ x gives −\(\frac{\pi}{2}<\theta<0.\) Now, sin \(\theta=\frac{x}{1}\) gives tan \(\theta=\frac{x}{\sqrt{1-x^2}}\)
In this case also, tan (sin^-1x) = \(\frac{x}{\sqrt{1-x^2}}\) ........(3)
Equations (1), (2) and (3) establish that tan (sin⁻¹x) \(=\frac{x}{\sqrt{1-x^{2}}},-1