Given, \(\frac { 4-\sqrt { 7 } }{ 3 } =tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 3 }{ 4 } \right) \right] \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 4-\sqrt { 7 } }{ 3 } \right) =\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 3 }{ 4 } \right) \right] \)
\(\Rightarrow 2{ tan }^{ -1 }\left( \frac { 4-\sqrt { 7 } }{ 3 } \right) ={ sin }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
Now,  L.H.S. \(=2{ tan }^{ -1 }\left( \frac { 4-\sqrt { 7 } }{ 3 } \right) \)
\(={ tan }^{ -1 }\left[ \frac { 2\left( \frac { 4-\sqrt { 7 } }{ 3 } \right) }{ { 1-\left( \frac { 4-\sqrt { 7 } }{ 3 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \left( \frac { 8-2\sqrt { 7 } }{ 3 } \right) }{ 1-\left( \frac { 16+7-8\sqrt { 7 } }{ 9 } \right) } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \left( \frac { 8-2\sqrt { 7 } }{ 3 } \right) }{ \frac { 9-23+8\sqrt { 7 } }{ 9 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 3(8-2\sqrt { 7 } ) }{ 8\sqrt { 7 } -14 } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 24-6\sqrt { 7 } }{ 8\sqrt { 7 } -14 } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 12-3\sqrt { 7 } }{ 4\sqrt { 7 } -7 } \times \frac { (-7-4\sqrt { 7 } }{ (-7-4\sqrt { 7 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { -84+84-48\sqrt { 7 } +21\sqrt { 7 } }{ 49-112 } \right] \)
\(={ tan }^{ -1 }\left[ \frac { -27\sqrt { 7 } }{ -63 } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 27\sqrt { 7 } }{ 63 } \right] \)
\(={ tan }^{ -1 }\left( \frac { 3 }{ \sqrt { 7 } } \right) \)
\(={ sin }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
= RHS

\(\tan ^{ -1 }{ 3x } +\tan ^{ -1 }{ 2x } =\frac { \pi }{ 4 } \)
\(\Rightarrow \tan ^{ -1 }{ \left( \frac { 3x+2x }{ 1-6{ x }^{ 2 } } \right) } =\frac { \pi }{ 4 } \)
\(\Rightarrow \frac { 5x }{ 1-6{ x }^{ 2 } } =1\)
\(\Rightarrow 6{ x }^{ 2 }+5x-1=0\)
\(\Rightarrow (6x-1)(x+1)=0\)
\(\Rightarrow x=\frac { 1 }{ 6 } ,x=-1\)
But on putting x = -1
The value of \(\tan ^{ -1 }{ 3x } +\tan ^{ -1 }{ 2x } \)  becomes-ve. Hence x = -1 is not possible.
Thus \(x=\frac {1}{6}\)
In real life, both type of situations occurred +ve and -ve, but it must be verified and then final result should be accepted. We should always be prepared to handle both the situations.

Firstly, let us assume
\(\tan ^{ -1 }{ a } =\alpha \Rightarrow \tan { \alpha } =a\)
\(\tan ^{ -1 }{ b } =\beta \Rightarrow \tan { \beta } =b\)
\(\tan ^{ -1 }{ c } =\gamma \Rightarrow \tan { \gamma } =c\)
Now, given that
\(\tan ^{ -1 }{ a } +\tan ^{ -1 }{ b } +\tan ^{ -1 }{ c } =\pi \)
\(\Rightarrow \alpha +\beta +\gamma =\pi \)
\(\Rightarrow \alpha +\beta =\pi -\gamma \)
Taking tangent on both sides, we have
\(\tan { \alpha +\beta } =\tan { \pi -\gamma } \)
\(\Rightarrow \frac { \tan { \alpha } +\tan { \beta } }{ 1-\tan { \alpha } .\tan { \beta } } =\tan { \pi -\gamma } \)
\(\Rightarrow \tan { \alpha } +\tan { \beta } =-\tan { \gamma } (1-\tan { \alpha } .\tan { \beta } )\)
\(\Rightarrow \tan { \alpha } +\tan { \beta } =-\tan { \gamma } +\tan { \alpha } .\tan { \beta } .\tan { \gamma } \)
Thus, a + b + c = abc

Given equation exists, if
x² + x \(\ge\) 0 and 0 < \(\sqrt { { x }^{ 2 }+x+1 } \le 1\)
[\(\because\)x² + x + 1 is always greater than zero]
Now, x² + x \(\ge\)0 and x² + x + 1 \(\le\)1
\(\Rightarrow\) x² + x \(\ge\)0 and x² + x \(\le\)0
\(\Rightarrow\) x² + x = 0 i.e., x(x + 1) = 0
Hence, x = 0 and -1 are the solution of the given equation.

\({ tan }^{ -1 }x+2{ cot }^{ -1 }x=\frac { 2\pi }{ 3 } \)
\(\Rightarrow \quad { tan }^{ -1 }x+2\left( \frac { \pi }{ 2 } -{ tan }^{ -1 }x \right) =\frac { 2\pi }{ 3 } \)
\(\Rightarrow \quad -{ tan }^{ -1 }x=\frac { 2\pi }{ 3 } -\pi \)
\(\Rightarrow \quad -{ tan }^{ -1 }x=-\frac { \pi }{ 3 } \)
\(\Rightarrow \quad { tan }^{ -1 }x=\frac { \pi }{ 3 } \)
\(\Rightarrow \quad x=tan\frac { \pi }{ 3 } \)
\(=\sqrt { 3 } \)

L.H.S = \(2{ sin }^{ -1 }\frac { 3 }{ 5 } -{ tan }^{ -1 }\frac { 17 }{ 31 } \)
\(=2{ tan }^{ -1 }\frac { 3 }{ 4 } -{ tan }^{ -1 }\frac { 17 }{ 31 } \)
\(={ tan }^{ -1 }\left( \frac { 2\left( \frac { 3 }{ 4 } \right) }{ 1-\frac { 9 }{ 16 } } \right) -{ tan }^{ -1 }\frac { 17 }{ 31 } \)
\(={ tan }^{ -1 }\left( \frac { 24 }{ 7 } \right) -{ tan }^{ -1 }\frac { 17 }{ 31 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 24 }{ 7 } -\frac { 17 }{ 31 } }{ 1+\frac { 24 }{ 7 } .\frac { 17 }{ 31 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 744-119 }{ 217+408 } \right) \)
\(={ tan }^{ -1 }(1)\\ =\frac { \pi }{ 4 } \)

Given that sin[ cot ^-1(x + 1)] = cos(tan^-1x)....(i)
We know that,
\({ cot }^{ -1 }(A)={ sin }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x + 1
Applying this identity in equation (i), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos({ tan }^{ -1 }x)\) ...(ii)
Also, we know that
\({ tan }^{ -1 }A={ cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x
Applying this identity in equation (ii), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos\left( { cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
\(\Rightarrow \frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } =\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \)
\(\left[ \because { sin }^{ -1 }(sin\theta )=\theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
and cos^-1 (cos \(\theta \)) = \(\theta \) \(\forall \theta \) [0, \(\pi \)]]
Squaring and Reciprocating both side, we have
1+(1 + x)² = 1 + x²
\(\Rightarrow\) 1 + 1 + x² + 2x = 1 + x²
\(\Rightarrow\) 1 + 2x = 0
\(\Rightarrow\)  \(x=-\frac { 1 }{ 2 } \quad \)

\(2\tan ^{ -1 }{ \frac { 1 }{ 2 } } =2\tan ^{ -1 }{ \frac { 2.\frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } } \)
\(=\tan ^{ -1 }{ \frac { 4 }{ 3 } } \)
\(LHS=\tan ^{ -1 }{ \frac { 4 }{ 3 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } } \)
\(=\tan ^{ -1 }{ \frac { \frac { 4 }{ 3 } +\frac { 1 }{ 7 } }{ 1-\frac { 4 }{ 3 } .\frac { 1 }{ 7 } } } \)
\(=\tan ^{ -1 }{ \frac { 31 }{ 17 } } =RHS\)

\({ sin }^{ -1 }\frac { 5 }{ 13 } ={ tan }^{ -1 }\frac { 5 }{ 12 } \)
\(and\quad { cos }^{ -1 }\frac { 3 }{ 5 } ={ tan }^{ -1 }\frac { 4 }{ 3 } \)
\(R.H.S.={ sin }^{ -1 }\frac { 5 }{ 13 } +{ cos }^{ -1 }\frac { 3 }{ 5 } \)
\(\\ ={ tan }^{ -1 }\frac { 5 }{ 12 } +{ tan }^{ -1 }\frac { 4 }{ 3 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 5 }{ 12 } +\frac { 4 }{ 3 } }{ 1-\frac { 5 }{ 12 } .\frac { 4 }{ 3 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =L.H.S.\)
Hence Proved

Getting,  \(\cos ^{ -1 }{ \frac { 12 }{ 13 } } =\tan ^{ -1 }{ \frac { 5 }{ 12 } } \)
\(\sin ^{ -1 }{ \frac { 3 }{ 5 } } =\tan ^{ -1 }{ \frac { 3 }{ 4 } } \)
and  \(\sin ^{ -1 }{ \frac { 56 }{ 65 } } =\tan ^{ -1 }{ \frac { 56 }{ 33 } } \)
\(L.H.S=\tan ^{ -1 }{ \frac { 5 }{ 12 } } +\tan ^{ -1 }{ \frac { 3 }{ 4 } } \)
\(=\tan ^{ -1 }{ \left[ \frac { \frac { 5+9 }{ 12 } }{ 1-\frac { 5 }{ 16 } } \right] } \)
\(=\tan ^{ -1 }{ \left( \frac { 14 }{ 12 } \times \frac { 16 }{ 11 } \right) } \)
\(=\tan ^{ -1 }{ \frac { 56 }{ 33 } } \)
\(=\sin ^{ -1 }{ \left( \frac { 56 }{ 33 } \right) } \)
\(=R.H.S\)

\(L.H.S.={ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 1 }{ 2 } +\frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 2 } \times \frac { 1 }{ 5 } } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 7 }{ 10 } }{ \frac { 9 }{ 10 } } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \)
\(={ tan }^{ -1 }\left( \frac { 7 }{ 9 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) ={ tan }^{ -1 }\left( \frac { \frac { 7 }{ 9 } +\frac { 1 }{ 8 } }{ 1-\frac { 7 }{ 9 } \times \frac { 1 }{ 8 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 65 }{ 72 } }{ \frac { 65 }{ 72 } } \right) \)
\(={ tan }^{ -1 }(1)\)
\(=\frac { \pi }{ 4 } =R.H.S.\)

\(\sin ^{ -1 }{ \left( \frac { 3 }{ 2 } \right) } =\tan ^{ -1 }{ \left( \frac { 2 }{ 3 } \right) } \)
\(\therefore \ \cos { \left[ \tan ^{ -1 }{ \left( \frac { 3 }{ 4 } \right) } +\tan ^{ -1 }{ \left( \frac { 2 }{ 3 } \right) } \right] } \)
\(=\cos { \left[ \tan ^{ -1 }{ \left( \frac { \frac { 3 }{ 4 } +\frac { 2 }{ 3 } }{ 1-\frac { 3 }{ 4 } \times \frac { 2 }{ 3 } } \right) } \right] } \)
\(=\cos { \left[ \tan ^{ -1 }{ \left( \frac { 17 }{ 6 } \right) } \right] } \)
\(=\cos { \left[ \cos ^{ -1 }{ \left( \frac { 6 }{ 5\sqrt { 13 } } \right) } \right] } \)
\(=\frac { 6 }{ 5\sqrt { 13 } } =RHS\)

\(\tan ^{ -1 }{ \left( \frac { \cos { x } }{ 1+\sin { x } } \right) } =\tan ^{ -1 }{ \left( \frac { \sin { \left( \frac { \pi }{ 2 } -x \right) } }{ 1+\cos { \left( \frac { \pi }{ 2 } -x \right) } } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 2\sin { \left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) } }{ \cos ^{ 2 }{ \left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) } } \right) } \)
\(=\tan ^{ -1 }{ \left( \tan { \left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) } \right) } \)
\(=\frac { \pi }{ 4 } -\frac { x }{ 2 } \)
Alternative Method:
LHS = \(\tan ^{ -1 }{ \left( \frac { \cos { x } }{ 1+\sin { x } } \right) } \)
\(\tan ^{ -1 }{ \left( \frac { \cos { x } }{ 1+\sin { x } } \right) } =\tan ^{ -1 }{ \left( \frac { \cos ^{ 2 }{ \frac { x }{ 2 } } \sin ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } +2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { \left( \sin { \frac { x }{ 2 } } -\cos { \frac { x }{ 2 } } \right) \left( \sin { \frac { x }{ 2 } } +\cos { \frac { x }{ 2 } } \right) }{ { \left( \sin { \frac { x }{ 2 } } +\cos { \frac { x }{ 2 } } \right) }^{ 2 } } \right) } \)
\(\left[ \therefore \quad x\epsilon \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \Rightarrow \frac { x }{ 2 } \epsilon \left( -\frac { \pi }{ 4 } ,\frac { \pi }{ 4 } \right) \right] \)
\(=\tan ^{ -1 }{ \left[ \frac { \left( \sin { \frac { x }{ 2 } } -\cos { \frac { x }{ 2 } } \right) }{ \left( \sin { \frac { x }{ 2 } } +\cos { \frac { x }{ 2 } } \right) } \right] } \)
\(=\tan ^{ -1 }{ \left( \frac { \left( 1-\tan { \frac { x }{ 2 } } \right) }{ \left( 1+\tan { \frac { x }{ 2 } } \right) } \right) } =\tan ^{ -1 }{ \left[ \frac { \tan { \frac { \pi }{ 4 } } -\tan { \frac { x }{ 2 } } }{ 1+\tan { \frac { \pi }{ 4 } } \tan { \frac { x }{ 2 } } } \right] } \)
\(=\tan ^{ -1 }{ \left[ \tan { \left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) } \right] } \)
\(=\frac { \pi }{ 4 } -\frac { x }{ 2 } \left[ \because \tan ^{ -1 }{ (\tan { \theta } ) } =\theta \forall \theta \epsilon \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
\(\therefore \ \tan ^{ -1 }{ \left( \frac { \cos { x } }{ 1+\sin { x } } \right) } =\frac { \pi }{ 4 } -\frac { x }{ 2 } =RHS\)

\(\tan ^{ -1 }{ 3x } +\tan ^{ -1 }{ 2x } =\frac { \pi }{ 4 } \)
\(\Rightarrow \tan ^{ -1 }{ \left( \frac { 3x+2x }{ 1-(3x)(2x) } \right) } =\frac { \pi }{ 4 } \)
\(\Rightarrow \frac { 5x }{ 1-{ 6x }^{ 2 } } =\tan { \frac { \pi }{ 4 } } \)
\(\Rightarrow \frac { 5x }{ 1-{ 6x }^{ 2 } } =1\)
\(\Rightarrow 5x=1-{ 6x }^{ 2 }\)
\(\Rightarrow\) 6x² + 5x - 1 = 0
\(\Rightarrow\) 6x² + 6x - x - 1 = 0
\(\Rightarrow\) 6x(x + 1) -1(x + 1) = 0
\(\Rightarrow\)(6x - 1) (x + 1) = 0
\(x=-1\quad or\quad \frac { 1 }{ 6 } \)

Given
\( \tan ^{-1}\left(\frac{x-2}{x-4}\right)+\tan ^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\left(\frac{x-2}{x-4}\right)\left(\frac{x+2}{x+4}\right)}\right]=\frac{\pi}{4} \) 
\(\Rightarrow \tan ^{-1}\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ {\left[\frac{(x-2)(x+4)+(x+2)(x-4)}{\frac{(x-4)(x+4)}{(x-4)(x+4)-(x-2)(x+2)}}{(x-4)(x+4)}\right]} \end{array}\right]=\frac{\pi}{4} \)
\(\Rightarrow \frac{x^{2}-2 x+4 x-8+x^{2}-4 x+2 x-8}{\left(x^{2}-16\right)-\left(x^{2}-4\right)}=\tan \frac{\pi}{4} \) 
\( \Rightarrow \frac{x^{2}-2 x+4 x-8+x^{2}-4 x+2 x-8}{\left(x^{2}-16\right)-\left(x^{2}-4\right)}=\tan \frac{\pi}{4} \)
\(\Rightarrow \frac{2 x^{2}-16}{-12}=1 \)
\(\Rightarrow 2 x^{2}-16=-12 \)
\(\Rightarrow 2 x^{2}=-12+16 \)
\(\Rightarrow 2 x^{2}=4 \Rightarrow x^{2}=2 \ \therefore \ x=\pm \sqrt{2}\)
\(\text {Hence, } \sqrt{2} \text { and }-\sqrt{2} \text { are the required value } \)

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

\(\text { We have } \tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)\)
\(\Rightarrow \tan ^{-1}\left(\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\left(\frac{x+1}{x-1}\right) \cdot\left(\frac{x-1}{x}\right)}\right)=\tan ^{-1}(-7) \)
\(\Rightarrow \tan ^{-1}\left(\frac{x(x+1)+(x-1)^{2}}{x(x-1)-\left(x^{2}-1\right)}\right)=\tan ^{-1}(-7) \)
\(\Rightarrow \frac{x^{2}+x+x^{2}-2 x+1}{x^{2}-x-x^{2}+1}=-7 \)
\(\Rightarrow \frac{2 x^{2}-x+1}{1-x}=-7 \)
\(\Rightarrow x^{2}-4 x+4=0 \Rightarrow(x-2)^{2}=0 \Rightarrow x=2 \)

\( \text { We have, } \tan \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right) \)
\(=\tan \frac{1}{2}\left(2 \tan ^{-1} x+2 \tan ^{-1} y\right) \) 
 \(\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right] \)
\(= \tan \frac{1}{2} \times 2\left(\tan ^{-1} x+\tan ^{-1} y\right) \)
\(= \tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \)
\(= \frac{x+y}{1-x y} \because \tan ^{-1} x+\tan ^{-1} y=\mid \tan ^{-1}\left(\frac{x+y}{1-x y}\right) \)
\(\text { and } \tan \left(\tan ^{-1} \theta\right)=\theta \)

\(\Rightarrow \tan ^{-1}\left(\frac{3 x+2 x}{1-3 x \times 2 x}\right)=\frac{\pi}{4}\)
\(\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right),\right. if \left.x y<1\right]\)
\(\Rightarrow\)\(\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\frac{\pi}{4}\)
\(\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4}\)
\(\left[\because \tan ^{-1}(\theta)=\phi \Rightarrow \theta=\tan \phi\right]\)
\(\Rightarrow \frac{5 x}{1-6 x^{2}}=1\)
\(\Rightarrow\)\(5 x=1-6 x^{2}\)
\(\Rightarrow 6 x^{2}+6 x-x-1=0\) 
\(\Rightarrow 6 x(x+1)-1(x+1)=0\)
\(\Rightarrow(6 x-1)(x+1)=0\)
\( 6 x-1=0 \Rightarrow x=\frac{1}{6}\)
and \(x+1=0 \Rightarrow x=-1\)
Hence, the required value of \(x \text { is } \frac{1}{6}\)

\( \text { We have, } \tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{8}{31}\)
\(\Rightarrow \tan ^{-1}\left(\frac{x+1+x-1}{1-(x+1)(x-1)}\right)=\tan ^ \frac{8}{31} \)
\(\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), \text { if } x y<1\right] \) 
\(\Rightarrow \tan ^{-1}\left(\frac{2 x}{1-x^{2}+1}\right)=\tan ^{-1} \frac{8}{31}\)
\(\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31} \Rightarrow 8\left(2-x^{2}\right)=62 x\)
\(\Rightarrow\)\(16-8 x^{2}-62 x=0\)
\(\Rightarrow\)\(-2\left(4 x^{2}+31 x-8\right)=0\)
\(\Rightarrow\)\(4 x^{2}+31 x-8=0\)
\(\Rightarrow 4 x^{2}+32 x-x-8=0\)
\(\Rightarrow 4 x(x+8)-1(x+8)=0\)
\(\Rightarrow (4 x-1)(x+8)=0\)
\(\Rightarrow 4 x-1=0\) or \(x+8=0\)
\(\Rightarrow x=\frac{1}{4}\) or \(x=-8\)
\(\therefore x=\frac{1}{4}\)

\( \text {Given, } \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right) \)
\(=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right) ;\left[\because \cos \frac{\pi}{3}=\frac{1}{2}, \sin \frac{\pi}{6}=\frac{1}{2}\right] \)
\(=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \)
\(=\frac{\pi}{3}+\frac{\pi}{3} \)
\(=\frac{2 \pi}{3} \)

Let \({ cosec }^{ -1 }\left( -\sqrt { 2 } \right) =y\), where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] -\left\{ 0 \right\} \)
\(\Rightarrow cosecy=-\sqrt { 2 } =-cosec\frac { \pi }{ 4 } \)
\(=cosec\left( -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

Let \({ cos }^{ -1 }\left( -\frac { 1 }{ \sqrt { 2 } } \right) =y\), where  \(y\in [0,\pi ]\)
\(\Rightarrow cosy=\frac { 1 }{ \sqrt { 2 } } \)
\(\Rightarrow cosy=-cos\frac { \pi }{ 4 } =cos\left( \pi -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow cosy=cos\frac { 3\pi }{ 4 } \Rightarrow y=\frac { 3\pi }{ 4 } \)
Hence, the required principal value = \(\frac { 3\pi }{ 4 } \)

Let \({ cot }^{ -1 }\left( \sqrt { 3 } \right) =y,0
\(\Rightarrow coty=\sqrt { 3 } \Rightarrow y=\frac { \pi }{ 6 } \)
Hence, the required principal value  \(\frac { \pi }{ 6 } \)

Let's consider \(\tan ^{-1}(1)=x\). Then, \(\tan x=1=\tan \left(\frac{\pi}{4}\right)\). \(\therefore \tan ^{-1}(1)=\frac{\pi}{4}\)
Let's assume,\(\cos ^{-1}\left(-\frac{1}{2}\right)=y\).
Then, \(\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)\)
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)
Let's again assume that \(\sin ^{-1}\left(-\frac{1}{2}\right)=z\).
Then, \(\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)\).
\(\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)
\(\therefore \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)
\(=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \)
\(=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\)

Let tan^-1(-1) = y, where \(y\in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(\Rightarrow tany=-1=-tan\frac { \pi }{ 4 } =tan\left( -\frac { \pi }{ 4 } \right) \)
\( \Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

Let tan⁻¹(−√3) = θ ⇒ tan θ = −√3
We know that the range of principal value of tan⁻¹ θ is (−π/2,π/2).
∴ tanθ = −√3=−tan π/3=tan(−π/3) (∵ tan(−θ) = −tanθ)
⇒ θ =−π/3, where θ ϵ (−π/2,π/2) ⇒tan⁻¹(−√3)=−π/3
Hence, principal value of tan−1(−√3) is −π/3

Let \({ tan }^{ -1 }\left( -\sqrt { 3 } \right) =y\) where \(y\in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(\Rightarrow tany=-\sqrt { 3 } =-tan\left( \frac { \pi }{ 3 } \right) \)
\(=tan\left( -\frac { \pi }{ 3 } \right)\)
\(\Rightarrow \ y=-\frac { \pi }{ 3 } \)
Hence, the reqd. principal value = \(-\frac { \pi }{ 3 } \)

Let \({ cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =y\), where \(y\in [0,\pi ]\)
\(\Rightarrow cosy=\frac { \sqrt { 3 } }{ 2 } \Rightarrow cosy=cos\frac { \pi }{ 6 } \)
\(\Rightarrow \ y=\frac { \pi }{ 6 } \)
Hence, the required Principal value = \(\frac { \pi }{ 6 } \)

Let \({ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) =y\) where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow siny=-\frac { \pi }{ 2 } \)
\(\Rightarrow siny=-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 6 } \)
Hence, the required principal value = \(-\frac { \pi }{ 6 } \)

We have 
\(\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]=\tan ^{-1}\left[\frac{\frac{a \cos x-b \sin x}{b \cos x}}{\frac{b \cos x+a \sin x}{b \cos x}}\right]=\tan ^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right]\)
\(=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)=\tan ^{-1} \frac{a}{b}-x\)

\(\text {Let } \sin ^{-1} \frac{12}{13}=x, \quad \cos ^{-1} \frac{4}{5}=y, \tan ^{-1} \frac{63}{16}=z\)
\( \sin x=\frac{12}{13}, \quad \cos y=\frac{4}{5}, \quad \tan z=\frac{63}{16}\)
\(\cos x=\frac{5}{13}, \sin y=\frac{3}{5}, \tan x=\frac{12}{5} \text { and } \tan y=\frac{3}{4}\)
\(\text {We have }\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}=-\frac{63}{16}\)
Hence tan(x + y) = − tan z
i.e., tan (x + y) = tan (–z) or tan (x + y) = tan (p – z)
Therefore x + y = – z or x + y = p – z
Since x, y and z are positive, x + y \(\ne\) – z (Why?)
\(\text {Hence }x+y+z=\pi \text { or } \sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi\)

\(\text {Let } \sin ^{-1} \frac{3}{5}=x \text { and } \sin ^{-1} \frac{8}{17}=y\)
\( \sin x=\frac{3}{5} \text { and } \sin y=\frac{8}{17}\)
\(\text {Now }\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
\(\text {and }\cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{64}{289}}=\frac{15}{17}\)
We have cos(x−y) = cos x cos y + sin x siny
\(=\frac{4}{5} \times \frac{15}{17}+\frac{3}{5} \times \frac{8}{17}=\frac{84}{85}\)
\( x-y=\cos ^{-1} \frac{84}{85}\)
\(\text {Hence } \ \sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}\)

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 \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)=\frac{2 \pi}{5}\)

Put x = tan \(\theta\) so that \(\theta\) = tan^-1x.
\(RHS={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ={ tan }^{ -1 }\left( \frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \right) \)
\(={ tan }^{ -1 }(tan3\theta )=3\theta =3{ tan }^{ -1 }x\)
\( ={ tan }^{ -1 }x+2{ tan }^{ -1 }x={ tan }^{ -1 }x+{ tan }^{ -1 }\frac { 2x }{ { 1-x }^{ 2 } } =LHS.\)

Let x = sec θ, then \(\sqrt{x^2-1}=\sqrt{\sec ^2 \theta-1}=\tan \theta\)
Therefore, \(\cot ^{-1} \frac{1}{\sqrt{x^2-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x\) which is the simplest form

First split \(\frac{2 \pi}{3} \text { as } \frac{(3 \pi-\pi)}{3} \text { or } \pi-\frac{\pi}{3}\)
After substuting in the given we get,
\(\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)=\frac{\pi}{3}\)

We write
\( \tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)= \tan ^{-1}\left[\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right] \)
\(= \tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}\right] \)
\(= \tan ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]=\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right] \)
\( =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2} \)

\(LHS={ tan }^{ -1 }\frac { 1 }{ 2 } +{ tan }^{ -1 }\frac { 2 }{ 11 } \)
\(={ tan }^{ -1 }\frac { \frac { 1 }{ 2 } +\frac { 2 }{ 11 } }{ 1-\frac { 1 }{ 2 } .\frac { 2 }{ 11 } } ={ tan }^{ -1 }\frac { 15 }{ 20 } ={ tan }^{ -1 }\frac { 3 }{ 4 } =RHS.\)

Take x = cos θ, then proceeding as above, we get, \({ \sin }^{ -1 }(2x\sqrt { 1-{ x }^{ 2 } } )\) = 2 cos^–1 x

Let sin^-1 x = \(\theta\) then  sin^-1 x = \(\theta\). we have
sin^-1 \((2x\sqrt { 1-{ x }^{ 2 } } )\) = sin^-1\(2\sin { \theta } \sqrt { 1-{ sin }^{ 2 } } \)
= sin^–1 (2sinθ cosθ) = sin^–1 (sin2θ) = 2θ = 2 sin^–1 x

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 \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) is \ \frac { \pi }{ 4 } \)

\(Let\quad { sec }^{ -1 }x=\theta ,\quad then\quad x=sec\theta \quad and\quad for\quad x<-1,\)
\(\frac { \pi }{ 2 } <\theta <\pi \)
Given expression = cot^-1(-cot \(\theta \))
\(={ cot }^{ -1 }\left[ cot\left( \pi -\theta \right) \right] =\pi -{ sec }^{ -1 }x\quad as\quad 0<\pi -\theta <\frac { \pi }{ 2 } \)

\({ sin }^{ -1 }\frac { 2a }{ 1+{ a }^{ 2 } } =2{ tan }^{ -1 }a\)
\(and\quad { sin }^{ -1 }\frac { 2b }{ 1+{ b }^{ 2 } } =2{ tan }^{ -1 }b\)
\(as\left[ 2{ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \right] \)
\(2{ tan }^{ -1 }a+2{ tan }^{ -1 }b=2ta{ n }^{ -1 }x\)
\({ tan }^{ -1 }a+{ tan }^{ -1 }b={ tan }^{ -1 }x\)
\({ tan }^{ -1 }\left( \frac { a+b }{ 1-ab } \right) =ta{ n }^{ -1 }x\)
\(x=\frac { a+b }{ 1-ab } \)  
Hence Proved.

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

\({ sin }^{ -1 }\left[ \frac { x+\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 2 } } \right] \quad Let\quad x=sin\theta \Rightarrow \theta ={ sin }^{ -1 }x\)
\(={ sin }^{ -1 }\left( \frac { sin\quad \theta +\sqrt { 1-{ sin }^{ 2 } } \theta }{ \sqrt { 2 } } \right) \)
\(={ sin }^{ -1 }\left( \frac { sin\theta +cos\theta }{ \sqrt { 2 } } \right) \)
\(={ sin }^{ -1 }\left( sin\theta \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } cos\theta \right) \)
\(={ sin }^{ -1 }\left( sin\theta cos\frac { \pi }{ 4 } +cos\theta sin\frac { \pi }{ 4 } \right) \)
\(={ sin }^{ -1 }\left[ \theta +\frac { \pi }{ 4 } \right] \)
\(\Rightarrow \theta +\frac { \pi }{ 4 } =\frac { \pi }{ 4 } +{ sin }^{ -1 }x\)

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

\(R.H.S.=2{ sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } \quad \quad \begin{cases} Let\quad x=cos\theta \\ \Rightarrow \theta ={ cos }^{ -1 }x \end{cases}\)
\(=2{ sin }^{ -1 }\left( \sqrt { \frac { 1-cos\theta }{ 2 } } \right) \)
\( \begin{cases} as\quad cos\theta =1=2{ sin }^{ 2 }\frac { \theta }{ 2 } \\ \Rightarrow 1-cos\theta =2sin^{ 2 }\frac { \theta }{ 2 } \end{cases}\)
\(=2{ sin }^{ -1 }\left( sin\frac { \theta }{ 2 } \right) \)
\(=2\left( \frac { \theta }{ 2 } \right) =\theta ={ cos }^{ -1 }x\)
L.H.S = R.H.S

We have \(x\in [-1,\quad 1]\)
Let  x = sin \(\theta\)             \(\therefore \quad \theta \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow \quad \quad \theta ={ sin }^{ -1 }x\)
\(\Rightarrow \quad \quad -\frac { \pi }{ 2 } \le \theta \le \frac { \pi }{ 2 } \)
\(\Rightarrow \quad \quad \frac { \pi }{ 2 } \ge -\theta \ge -\frac { \pi }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 2 } +\frac { \pi }{ 2 } \ge \frac { \pi }{ 2 } -\theta \ge -\frac { \pi }{ 2 } +\frac { \pi }{ 2 } \)
\(\Rightarrow \quad\pi \ge \frac { \pi }{ 2 } -\theta \ge 0\)
\(\Rightarrow \quad cos\left( \frac { \pi }{ 2 } -\theta \right) =sin\quad \theta =x\)
\(\Rightarrow \quad \frac { \pi }{ 2 } -\theta ={ cos }^{ -1 }x\)
\(\Rightarrow { \quad cos }^{ -1 }x=\frac { \pi }{ 2 } -{ sin }^{ -1 }x\)
\(\Rightarrow { \quad sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } if\quad x\in [-1,\quad 1]\)

\(LHS={ tan }^{ -1 }\frac { 1 }{ 4 } +{ tan }^{ -1 }\frac { 2 }{ 9 } \)
\(={ tan }^{ -1 }\left[ \frac { \frac { 1 }{ 4 } +\frac { 2 }{ 9 } }{ 1-\frac { 1\times 2 }{ 4\times 9 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \frac { 9+8 }{ 36 } }{ \frac { 36-2 }{ 36 } } \right] \)
\(={ tan }^{ -1 }\left( \frac { 17 }{ 34 } \right) ={ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=\frac { 1 }{ 2 } \left( 2{ tan }^{ -1 }\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 2\times \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } \right) \)
\(\because \left[ 2{ tan }^{ -1 }={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 1 }{ 3/4 } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)

\(L.H.S.={ tan }^{ -1 }\left( \frac { 3{ a }^{ 2 }x-{ x }^{ 3 } }{ { a }^{ 3 }-3a{ x }^{ 2 } } \right) \)
\(Divide\quad by\quad a^{ 3 },\)
\(={ tan }^{ -1 }\left( \frac { 3\frac { x }{ a } -\frac { { x }^{ 3 } }{ { a }^{ 3 } } }{ 1-3\left( \frac { { x }^{ 2 } }{ { a }^{ 2 } } \right) } \right) \)
\(={ tan }^{ -1 }\left[ \frac { 3\left( \frac { x }{ a } \right) -{ \left( \frac { x }{ a } \right) }^{ 3 } }{ 1-3{ \left( \frac { x }{ a } \right) }^{ 2 } } \right] \)
\(Put,\frac { x }{ a } =tan\theta ,\quad \theta ={ tan }^{ -1 }\frac { x }{ a } \)
\(={ tan }^{ -1 }\left[ \frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \right] \)
\(={ tan }^{ -1 }(tan\quad 3\theta )\)
\(3\theta =3{ tan }^{ -1 }\left( \frac { x }{ a } \right) =R.H.S.\)

\(L.H.S.={ tan }^{ -1 }\frac { 2 }{ 3 } =\frac { 1 }{ 2 } \left( 2{ tan }^{ -1 }\frac { 2 }{ 3 } \right) \)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }\left( \frac { 2\times \frac { 2 }{ 3 } }{ 1-\frac { 4 }{ 9 } } \right) \right] \)
\(\left[ \because 2{ tan }^{ -1 }x={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left[ \frac { \frac { 4 }{ 3 } }{ \frac { 9-4 }{ 9 } } \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 4\times 9 }{ 3\times 5 } \right) \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 12 }{ 5 } \right) =R.H.S.\)

\(4\quad { tan }^{ -1 }\frac { 1 }{ 5 } =2\left[ 2{ tan }^{ -1 }\frac { 1 }{ 5 } \right] \)
\(=2\left[ { tan }^{ -1 }\left( \frac { 2\times \frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 25 } } \right) \right] \)
\(\left[ \because 2{ tan }^{ -1 }x={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=2{ tan }^{ -1 }\left( \frac { \frac { 2 }{ 5 } }{ \frac { 25-1 }{ 25 } } \right) \)
\(=2{ tan }^{ -1 }\left( \frac { 2\times 25 }{ 24\times 5 } \right) =2{ tan }^{ -1 }\left( \frac { 5 }{ 12 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 2\times 5 }{ 12 } }{ 1-\frac { 25 }{ 144 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 10 }{ 12 } }{ \frac { 144-25 }{ 144 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 10\times 144 }{ 119\times 12 } \right) \)
\(={ tan }^{ -1 }\left( \frac { 120 }{ 119 } \right) \)

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

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

\({ tan }^{ -1 }\frac { 3 }{ 4 } +{ tan }^{ -1 }\frac { 3 }{ 5 } -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 15+12 }{ 20 } }{ \frac { 20-9 }{ 20 } } \right) -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(\left[ \because { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\frac { 27 }{ 11 } -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\left( \frac { \frac { 27 }{ 11 } -\frac { 8 }{ 19 } }{ 1+\frac { 27\times 8 }{ 11\times 19 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 513-88 }{ 209 } }{ \frac { 209+216 }{ 209 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 425 }{ 425 } \right) \)
\(={ tan }^{ -1 }1=\frac { \pi }{ 4 } \)

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