L.H.S = \( { tan }^{ -1 }\frac { 1 }{ 7 } +{ tan }^{ -1 }\frac { 1 }{ 8 } +{ tan }^{ -1 }\left( \frac { 1 }{ 18 } \right) \left( { cot }^{ -1 }\theta =tan\frac { 1 }{ \theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 1 }{ 7 } +\frac { 1 }{ 8 } }{ 1-\frac { 1 }{ 56 } } \right) +{ tan }^{ -1 }\frac { 1 }{ 18 } \)
\(={ tan }^{ -1 }\left( \frac { 3 }{ 11 } \right) +{ tan }^{ -1 }\frac { 1 }{ 18 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 3 }{ 11 } +\frac { 1 }{ 18 } }{ 1-\frac { 3 }{ 11 } \times \frac { 1 }{ 18 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 54+11 }{ 198-3 } \right) ={ tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) \)
\(={ cot }^{ -1 }(3)=RHS\)

Putting x = cos \(\theta \) in L.H.S., we get
\(LHS={ tan }^{ -1 }\left[ \frac { \sqrt { 1+cos\theta } -\sqrt { 1-cos\theta } }{ \sqrt { 1+cos\theta } +\sqrt { 1-cos\theta } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \sqrt { 2 } cos\frac { \theta }{ 2 } -\sqrt { 2 } sin\frac { \theta }{ 2 } }{ \sqrt { 2 } cos\frac { \theta }{ 2 } +\sqrt { 2 } sin\frac { \theta }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 1-tan\frac { \theta }{ 2 } }{ 1+tan\frac { \theta }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) \right] \)
\(=\frac { \pi }{ 4 } -\frac { \theta }{ 2 } \)
\(=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } { cos }^{ -1 }x=RHS\)

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.

LHS = \(\cos { \left[ \tan ^{ -1 }{ \left\{ \sin { (\cot ^{ -1 }{ x } ) } \right\} } \right] } \)
Let \(\cot ^{ -1 }{ x } =\theta \ \Rightarrow \ x=\cot { \theta } \)
Now,  LHS = \(\cos { \left[ \tan ^{ -1 }{ \left\{ \sin { (\sin { \theta } ) } \right\} } \right] } \)
\(=\cos { \left[ \tan ^{ -1 }{ \left\{ \frac { 1 }{ cosec\theta } \right\} } \right] } \)
\(\\ =\cos { \left[ \tan ^{ -1 }{ \frac { 1 }{ \sqrt { 1+\cot ^{ 2 }{ \theta } } } } \right] } \)
\(=\cos { \left[ \tan ^{ -1 }{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } \right] } \)
Let  \(\tan ^{ -1 }{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } =\alpha \quad \Rightarrow \quad \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } =\tan { \alpha } \)
\(\Rightarrow \quad \frac { 1 }{ 1+{ x }^{ 2 } } =\tan ^{ 2 }{ \alpha } \)
\(\Rightarrow \quad \frac { 1 }{ 1+{ x }^{ 2 } } =\frac { \sin ^{ 2 }{ \alpha } }{ \cos ^{ 2 }{ \alpha } } \)
\(\Rightarrow \quad \frac { 1 }{ 1+{ x }^{ 2 } } +1=\frac { \sin ^{ 2 }{ \alpha } }{ \cos ^{ 2 }{ \alpha } } +1\)
\(\Rightarrow \quad \frac { 2+{ x }^{ 2 } }{ 1+{ x }^{ 2 } } +1=\frac { 1 }{ \cos ^{ 2 }{ \alpha } } \quad \Rightarrow \quad \cos { \alpha } =\frac { \sqrt { 1+{ x }^{ 2 } } }{ \sqrt { 2+{ x }^{ 2 } } } \)
\(\Rightarrow \quad \alpha =\cos ^{ -1 }{ \left( \frac { \sqrt { 1+{ x }^{ 2 } } }{ \sqrt { 2+{ x }^{ 2 } } } \right) } \)
\(\Rightarrow \quad LHS=\cos { \left( \cos ^{ -1 }{ \sqrt { \frac { 1+{ x }^{ 2 } }{ 2+{ x }^{ 2 } } } } \right) } \)
\(=\quad \sqrt { \frac { 1+{ x }^{ 2 } }{ 2+{ x }^{ 2 } } } .\)

LHS = \(2\left[ { tan }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right] +{ sec }^{ -1 }\left( \frac { 5\sqrt { 2 } }{ 7 } \right) \)
\(=2{ tan }^{ -1 }\left( \frac { \frac { 1 }{ 5 } +\frac { 1 }{ 8 } }{ 1-\frac { 1 }{ 40 } } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(=2{ tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } }{ 1-\left( \frac { 1 }{ 3 } \right) ^{ 2 } } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\frac { \frac { 3 }{ 4 } +\frac { 1 }{ 7 } }{ 1-\frac { 3 }{ 28 } } ={ tan }^{ -1 }\left( \frac { 25 }{ 25 } \right) ={ tan }^{ -1 }(1)=\frac { \pi }{ 4 } 1\)

Let, \(\sin ^{ -1 }{ x } =\sin { A } \quad \Rightarrow \quad \sin { A } =x\)
\(\sin ^{ -1 }{ y } =\sin { B } \quad \Rightarrow \quad \sin { B } =y\)
\(\sin ^{ -1 }{ z } =\sin { C } \quad \Rightarrow \quad \sin { C } =z\)
\(\Rightarrow \quad A+B+C=\pi \quad \Rightarrow \quad 2A+2B+2C=2\pi \)
\(\therefore \quad sin2A+sin2B+sin2C=4sinAsinBsinC\)
[Using trigonometric property]
\(\Rightarrow \quad 2sinAcosA+2sinBcosB+2sinCcosC=4sinAsinBsinC\)
\(\Rightarrow \quad 2sinA\sqrt { 1-\sin ^{ 2 }{ A } } +2sinB\sqrt { 1-\sin ^{ 2 }{ B } } +2sinC\sqrt { 1-\sin ^{ 2 }{ C } } =4sinAsinBsinC\)
\(\Rightarrow \quad 2x\sqrt { 1-{ x }^{ 2 } } +2y\sqrt { 1-{ y }^{ 2 } } +2z\sqrt { 1-{ z }^{ 2 } } =4xyz\)
\(\Rightarrow \quad x\sqrt { 1-{ x }^{ 2 } } +y\sqrt { 1-{ y }^{ 2 } } +2z\sqrt { 1-{ z }^{ 2 } } =2xyz\)
Hence proved.

\(=2\tan ^{ -1 }{ \left\{ \tan { \frac { \alpha }{ 2 } } \tan { \left( \frac { \pi }{ 4 } -\frac { \beta }{ 2 } \right) } \right\} } \)
\(=\tan ^{ -1 }{ \frac { 2\tan { \frac { \alpha }{ 2 } } .\tan { \left( \frac { \pi }{ 4 } -\frac { \beta }{ 2 } \right) } }{ 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \tan ^{ 2 }{ \left( \frac { \pi }{ 4 } -\frac { \beta }{ 2 } \right) } } } \)   \(\left[ \because \quad 2\tan ^{ -1 }{ x } =\tan ^{ -1 }{ \frac { 2x }{ 1-{ 2x }^{ 2 } } } \right] \)
\(=\tan ^{ -1 }{ \frac { 2\tan { \frac { \alpha }{ 2 } } \left[ \frac { 1-\tan { \frac { \beta }{ 2 } } }{ 1+\tan { \frac { \beta }{ 2 } } } \right] }{ 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } { \left( \frac { 1-\tan { \frac { \beta }{ 2 } } }{ 1+\tan { \frac { \beta }{ 2 } } } \right) }^{ 2 } } } } \)
\(\left[ \because \tan { (A-B) } =\frac { \tan { A } -\tan { B } }{ 1+\tan { A } \tan { B } } \right] \)
\(=\tan ^{ -1 }{ \left( \frac { 2\tan { \frac { \alpha }{ 2 } } \left[ \frac { \left( 1-\tan { \frac { \beta }{ 2 } } \right) \left( 1+\tan { \frac { \beta }{ 2 } } \right) }{ { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } } \right] }{ \frac { { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 }-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } { \left( 1-\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } }{ { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } } } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 2\tan { \frac { \alpha }{ 2 } } .\left( 1-\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) }{ { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 }-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } { \left( 1-\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } } \right) } \)
\(=\tan ^{ -1 }{ \left[ \frac { 2\tan { \frac { \alpha }{ 2 } } .\left( 1-\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) }{ \left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) \left( 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) +2\tan { \frac { \beta }{ 2 } } \left( 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) } \right] } \)
\(=\tan ^{ -1 }{ \left[ \frac { \frac { 2\tan { \frac { \alpha }{ 2 } } }{ \left( 1+\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) } .\frac { \left( 1-\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) }{ \left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) } }{ \frac { 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } }{ \left( 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) } +\frac { 2\tan { \frac { \beta }{ 2 } } }{ \left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) } } \right] } \)
\(=\tan ^{ -1 }{ \frac { \sin { \alpha } +\cos { \beta } }{ \cos { \alpha } +\sin { \beta } } } \)
[Dividing N^, and D ^' by \(\left( 1+\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) \)\(\left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) \)]
= RHS

Let, \({ cos }^{ -1 }x=\alpha \Rightarrow x=cos\alpha \)
LHS=\(\alpha +{ cos }^{ -1 }\left[ cos\alpha cos\left( \frac { \pi }{ 3 } \right) +\frac { \sqrt { 3 } }{ 2 } \sqrt { 1-{ cos }^{ 2 }\alpha } \right] \)
\(=\alpha +{ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 3 } \right) cos\alpha +sin\frac { \pi }{ 3 } sin\alpha \right] \)
\(=\alpha +{ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 3 } -\alpha \right) \right] \)
\(=\alpha +\frac { \pi }{ 3 } -\alpha \)
\(\left[ \because \quad { cos }^{ -1 }(cos\theta )=\theta \forall \theta \in (0,\pi ) \right] \)
\(=\frac { \pi }{ 3 } =RHS\)

\({ cot }^{ -1 }\left( \frac { \sqrt { 1+sin\quad x } +\sqrt { 1-sin\quad x } }{ \sqrt { 1+sin\quad x } -\sqrt { 1-sin\quad x } } \right) \)
\(={ cot }^{ -1 }\left[ \frac { \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +{ sin }^{ 2 }\frac { x }{ 2 } +2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) + } }{ \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +{ sin }^{ 2 }\frac { x }{ 2 } +2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) - } } \frac { \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +{ sin }^{ 2 }\frac { x }{ 2 } -2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) } }{ \sqrt { \left( { cos }^{ 2 }\frac { x }{ 2 } +sin^{ 2 }\frac { x }{ 2 } -2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right) } } \right] \)
\(={ cot }^{ -1 }\left[ \frac { \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } +\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } }{ \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 }-{ \sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } } } } \right] \)
\(={ cot }^{ -1 }\left[ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) +\left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) -\left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) } \right] \)
\(={ cot }^{ -1 }\left[ \frac { 2cos\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } } \right] \left[ \because \quad { cot }^{ -1 }(cot\quad \theta )=\forall \theta \in (0,\quad N) \right] \)
\(={ cot }^{ -1 }\left[ cot\frac { x }{ 2 } \right] \)
\(=\frac { x }{ 2 } \)

\(\cos { (\tan ^{ -1 }{ x } ) } =\cos { \left[ \cos ^{ -1 }{ \frac { 1 }{ \sqrt { { x }^{ 2 }+1 } } } \right] } \)
\(=\frac { 1 }{ \sqrt { { x }^{ 2 }+1 } } \quad \left[ \because \cos { (\cos ^{ -1 }{ x } ) } =x\forall x\epsilon [-1,1] \right] \)
\(\cot ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }+1 } } \right) } =\sin ^{ -1 }{ \left( \frac { \sqrt { { x }^{ 2 }+1 } }{ \sqrt { { x }^{ 2 }+2 } } \right) } \)
\(\therefore \sin { \left[ \cot ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }+1 } } \right) } \right] } \)
\(=\sin { \left[ \sin ^{ -1 }{ \left( \frac { \sqrt { { x }^{ 2 }+1 } }{ \sqrt { { x }^{ 2 }+2 } } \right) } \right] } \)
\(=\frac { \sqrt { { x }^{ 2 }+1 } }{ \sqrt { { x }^{ 2 }+2 } } \)
\(\left[ \because \sin { (\sin ^{ -1 }{ x } ) } =x\forall x\epsilon [-1,1] \right] \)

\({ tan }^{ -1 }\left( \frac { \sqrt { 1+cos\quad x } +\sqrt { 1-cos\quad x } }{ \sqrt { 1+cos\quad x\quad } -\sqrt { 1-cos\quad x } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \sqrt { 2\quad { cos }^{ 2 }\frac { x }{ 2 } } +\sqrt { 2{ sin }^{ 2 }\frac { x }{ 2 } } }{ \sqrt { 2\quad { cos }^{ 2 }\frac { x }{ 2 } } -\sqrt { 2{ sin }^{ 2 }\frac { x }{ 2 } } } \right) \)
\(={ tan }^{ -1 }\left( \frac { -\sqrt { 2 } cos\frac { x }{ 2 } +\sqrt { 2 } sin\frac { x }{ 2 } }{ -\sqrt { 2 } cos\frac { x }{ 2 } -\sqrt { 2 } sin\frac { x }{ 2 } } \right) \)
\(\left( As\quad \pi 0 \right) \)
\(={ tan }^{ -1 }\left( \frac { 1-tan\frac { x }{ 2 } }{ 1+tan\frac { x }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) \right) \)
\(\\ =\frac { \pi }{ 4 } -\frac { x }{ 2 } \ \left( As-\frac { \pi }{ 4 } >\frac { \pi }{ 4 } -\frac { x }{ 2 } >-\frac { x }{ 2 } \right) \)

Given that
sin^-1x + sin^-1(1 - x) = cos^-1x
Taking sin of both sides,
sin(sin^-1x + sin^-1(1 - x)) = sin(cos^-1x)
\(\Rightarrow\) sin (sin^-1x)cos(sin^-1(1 - x)) + cos(sin^-1x) sin(sin^-1(1 - x)) = sin(cos^-1x)
\(\Rightarrow x\sqrt { 1-{ (1-x) }^{ 2 } } +(1-x)\sqrt { 1-{ x }^{ 2 } } \)
\(=\sqrt { 1-{ x }^{ 2 } } \)
\([\because sin^{ -1 }(sin\quad \theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(and\quad { cos }^{ -1 }(cos\theta )=\theta \forall \theta \in [0.\quad \pi ]]\)
\([Here\ cos\{ { sin }^{ -1 }(1-x)\} ]=\sqrt { 1-{ sin }^{ 2 }[{ sin }^{ -1 }(1-x)] } \)
\(=\sqrt { 1-[sin\{ { sin }^{ -1 }(1-x)\} ]^{ 2 } } \)
\(=\sqrt { { 1-(1-x) }^{ 2 } } \)
\(cos\ ({ sin }^{ -1 }x)=\sqrt { 1-{ sin }^{ 2 }({ sin }^{ -1 }x) } \)
\(=\sqrt { 1-{ x }^{ 2 } } \)
\(\Rightarrow x\sqrt { 2x-{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } (1-x-1)=0\)
\(\Rightarrow x\left( \sqrt { 2x-{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) =0\)
\(\Rightarrow x=0\quad or\quad 2x-{ x }^{ 2 }=1-{ x }^{ 2 }\)
\(\Rightarrow x=0\quad or\quad x=\frac { 1 }{ 2 } \) 

\(y=\cot ^{ -1 }{ (\sqrt { \cos { x } } ) } -\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } ,\)
\(\Rightarrow y=\frac { \pi }{ 2 } -\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } -\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } \)
\(\Rightarrow y=\frac { \pi }{ 2 } -2\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } \)
\(\Rightarrow y=\frac { \pi }{ 2 } -\left[ \frac { (2\sqrt { \cos { x } } ) }{ 1-{ (\sqrt { \cos { x } } ) }^{ 2 } } \right] \)
\(\Rightarrow \left[ \frac { (2\sqrt { \cos { x } } ) }{ 1-{ (\sqrt { \cos { x } } ) }^{ 2 } } \right] =\frac { \pi }{ 2 } -y\)
\(\Rightarrow \frac { (2\sqrt { \cos { x } } ) }{ 1-\cos { x } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ \sqrt { { (2\sqrt { \cos { x } } ) }^{ 2 } } +{ (1+\cos { x } ) }^{ 2 } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ \sqrt { 1+\cos ^{ 2 }{ x } +2\cos { x } } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ { \sqrt { (1+\cos { x } } ) }^{ 2 } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ 1+\cos { x } } =\frac { 2\sin ^{ 2 }{ \frac { x }{ 2 } } }{ 2\cos ^{ 2 }{ \frac { x }{ 2 } } } \)
\(\Rightarrow \sin { y } =\tan ^{ 2 }{ \frac { x }{ 2 } } \)

\(a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=\ldots=a_{n}-a_{n-1}=d\)
\(\tan \left[\tan ^{-1}\left(\frac{d}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{d}{1+a_{2} a_{3}}\right)+\tan ^{-1}\left(\frac{d}{1+a_{3} a_{4}}\right)+\ldots . .+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_{n}}\right)\right]\)
\(=\tan \left[\tan ^{-1} \frac{a_{2}-a_{1}}{1+a_{1} a_{2}}+\tan ^{-1} \frac{a_{3}-a_{2}}{1+a_{2} a_{3}}+\ldots .+\tan ^{-1} \frac{a_{n}-a_{n-1}}{1+a_{n-1} a_{n}}\right]\)
\(=\tan \left[\tan ^{-1} a_{2}-\tan ^{-1} a_{1}+\tan ^{-1} a_{3}-\tan ^{-1} a_{2}+\ldots . .+\tan ^{-1} a_{n}-\tan ^{-1} a_{n-1}\right]=\tan \left[\tan ^{-1} a_{n}-\tan ^{-1} a_{1}\right]\)
\(=\tan \left\{\tan ^{-1}\left(\frac{a_{n}-a_{1}}{1+a_{n} a_{1}}\right)\right\}=\frac{a_{n}-a_{1}}{1+a_{n} a_{1}}\)

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

\(\text {We have } 2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x) \)
\(\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}\left(\frac{2}{\cos x}\right) \)
\(\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=\frac{2}{\cos x} \Rightarrow \frac{\sin x}{\cos x}=1 \)
\(\Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \)

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