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

\({ tan }^{ -1 }\sqrt { x(x+1) } +{ sin }^{ -1 }\sqrt { { x }^{ 2 }+x+1 } =\frac { \pi }{ 2 } \)
This equation holds if
\({ x }^{ 2 }+x\ge 0\) and \({ x }^{ 2 }+x+1\le 1\)
Now, \({ x }^{ 2 }+x\le 0\) and \(0\le { x }^{ 2 }+x+1\le 1\)
\(\Rightarrow { x }^{ 2 }+x\ge 0\quad { x }^{ 2 }+x+1\le 1\) [\( \because { x }^{ 2 }+x+1\ge 0 \) for all x]
\( \Rightarrow x^{2}+x \geq 0 \text { and } x^{2}+x \leq 0 \)
\( \Rightarrow x^{2}+x=0 \Rightarrow x=0,-1 \)
Hence, x = 0, -1 are the solutions of the given equation.

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow cos\theta =\frac { 4 }{ 5 } \)
\(\therefore tan\theta =\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\(\Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
LHS = \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) +{ tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
\(\left[ \because { tan }^{ -1 }x+tan^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 3 }{ 4 } +\frac { 3 }{ 5 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 3 }{ 5 } \right) } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 15+12 }{ 20 } }{ \frac { 20-9 }{ 20 } } \right) ={ tan }^{ -1 }\left( \frac { 27 }{ 20 } \times \frac { 20 }{ 11 } \right) \)
=  \({ tan }^{ -1 }\left( \frac { 27 }{ 11 } \right) =RHS\)
Hence proved

\({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) +{ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) ={ cos }^{ -1 }\left( cos\left( 2\pi -\frac { 2\pi }{ 3 } \right) \right) \)
[\(\therefore\) the period of cosine in 2\(\pi \)]
= \({ cos }^{ -1 }\left( cos\frac { 2\pi }{ 3 } \right) \) \(\left[ \because cos\left( 2\pi -\theta \right) =cos\theta \right] \)
= \(\frac { 2\pi }{ 3 } \) \(\left[ \because \frac { 2\pi }{ 3 } \epsilon \left[ 0,\pi \right] \right] \)
\({ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) ={ cop }^{ -1 }\left( cos\left( 2\pi -\frac { 3\pi }{ 4 } \right) \right) \) = [ \(\therefore\) Period of cosine is 2\(\pi\)]
 \({ cos }^{ -1 }\left( cos\left( 3\frac { \pi }{ 4 } \right) \right) \)  \(\left[ \because cos\left( 2\pi -\theta \right) =cos\theta \right] \)
= \(3\frac { \pi }{ 4 } \) \(\left[ \because 3\frac { \pi }{ 4 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\therefore \) \({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) +{ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \(\frac { 2\pi }{ 3 } +3\frac { \pi }{ 4 } \)
= \(\frac { 8\pi +9\pi }{ 12 } =\frac { 7\pi }{ 1b } \)
\({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) +{ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) \) = \(\frac { 7\pi }{ 12 } \)

Given that sin^-1x > cos^-1x. Note that -1\(\le x\le\)
Adding both sides by sin^-1x, we get
sin^-1 x + sin^-1 x > x cos^-1 x + sin^-1x, Which resucess to 2 sin^-1 x > \(\frac{\pi}{2}\)
As sine function increases in the interval \(\left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \), we have x > sin\(\frac{\pi}{4} or x> \frac{1}{\sqrt2}\)
Thus, the solution set is the interval \(\left[ \frac { 1 }{ \sqrt { 2 } } ,1 \right] \)

Let sin⁻¹ x = \(\theta\).
Then, x = sin \(\theta\) and-1\(\le x\le1\)
Now, tan(sin^-1 x) = tan \(\theta\) = \(\frac{sin\theta}{cos\theta}=\frac{sin\theta}{\sqrt{1-sin^{2}\theta}}=\frac{x}{\sqrt{1-x2}},|x|<1\)

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\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

Let tan⁻¹(−1) = y. Then, tan y = -1 = -tan\(\frac{\pi}{4}=tan(-\frac{\pi}{4})\)
As -\(\frac{\pi}{4}\in (-\frac{\pi}{2},\frac{\pi}{2}), tan^-1(-1)=-\frac{\pi}{ 3}\)
Now, cos^-1\((\frac{1}{2})\) = y implies cos y = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
As \(\frac{\pi}{3}\)\(\in\)[0, \(\pi\)], cos^-1 \((\frac{1}{2})=\frac{\pi}{3}\)
Now, sin^-1\((-\frac{1}{2})\) = y implies sin y = -\(\frac{1}{2}\) = sin(-\(\frac{\pi}{3}\)).
As -\(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}], sin^-1(-\frac{1}{2})=-\frac{\pi}{6}\)
Therefore, tan⁻¹(−1)+cos^-1\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

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

Given \(f(x)=sin^{ -1 }\left( \frac { x^{ 2 }+1 }{ 2x } \right) \le 1\)
We know that the domain of sin^-1(x) is [-1, 1]
\(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \le 1\)
Consider \(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \)
\(\Rightarrow 0\le \frac { { x }^{ 2 }+1 }{ 2x } +1\)
\(\Rightarrow \frac { { x }^{ 2 }+1+2x }{ 2x } \ge 0\)
\(\Rightarrow \frac { \left( x+1 \right) ^{ 2 } }{ 2x } \ge 0\)
\(\Rightarrow \) x = -1 and x < 0
Consider \(\cfrac { { x }^{ 2 }+1 }{ 2x } \le 1\)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ 2x } -1\le 0\)
\(\Rightarrow \frac { { x }^{ 2 }-2x+1 }{ 2x } \le 0\)
\(\Rightarrow \frac { \left( x-1 \right) ^{ 2 } }{ 2x } \le 0\)
From (1) and (2) Domain {-1, 1}

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

LHS = \(2{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } +\cfrac { 2 }{ 3 } }{ 1-\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } \right) } \right) ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } }{ \frac { 9-2 }{ 9 } } \right) \)
= \({ tan }^{ -1 }\left( \frac { 4 }{ 3 } \times \frac { 9 }{ 7 } \right) ={ tan }^{ -1 }\left( \frac { 12 }{ 7 } \right) \)
= RHS
Hence proved

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\) where \(0\le y\le \pi \)
Then \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\Rightarrow cosy=\frac { -1 }{ 2 } \)
\(\Rightarrow cos\ y=-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\left( \frac { 2\pi }{ 3 } \right) \)
\(\Rightarrow y=\frac { 2\pi }{ 3 } \left[ \because \frac { 2\pi }{ 3 } \in \left[ 0,\pi \right] \right] \)
\(\therefore \) The principal value of \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) \frac { 2\pi }{ 3 } \)

Let sin-1(-1) = y where \(\frac { -\pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
Then \(sin^{ -1 }(-1)=y\Rightarrow sin\quad y=-1\)
\(-1=sin\left( \frac { -\pi }{ 2 } \right) \Rightarrow y=\frac { -\pi }{ 2 } \) \(\left[ \because \frac { -\pi }{ 2 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\( \therefore\)  The principal value of \({ sin }^{ -1 }(-1)\ is \ \frac { -\pi }{ 2 } \)

Let y = sec^-1 (-2). Then, sec y = -2
By the definition, the range of the principal value branch of y = sec⁻¹x is [0, \(\pi\)]\{\({{\frac{\pi}{2}}}\)}
Let us find y in [0, \(\pi\)] - {\({{\frac{\pi}{2}}}\)} such that sec y = -2
But, sec y = −2 \(\Rightarrow\) cos y = -\(\frac{1}{2}\)
Now, cos y = -\(\frac { 1 }{ 2 } =-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\frac { 2\pi }{ 3 } \). Therefore, y = \(\frac{2\pi}{3}\)
since \(\frac{2\pi}{3}\in[0,\pi]\)\{\({{\frac{\pi}{2}}}\)}, the principal value of sec^-1(-2) is \(\frac{2\pi}{3}\)

\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi +\frac { \pi }{ 4 } \right) \right) \) \(\because \frac { 5\pi }{ 4 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
= \({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 4 } \right) \right) \)
= \( \frac {- \pi }{ 4 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) 

y = 4 sin (-2x)
The amplitude of sin x is 1
\(\Rightarrow\) amplitude of sin (-2x) is 1
\(\therefore\) Amplitude 4 sin(-2x) is 4 \(\times\) 1 = 4.
The period of sin \((-2x)is2x=2\pi \Rightarrow =\frac { 2\pi }{ 2 } =\pi \)

Since tan(tan^-1 x) = x, x ∈ R,
We have tan(tan^-1(2019)) = 2019

It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1 and 0\le y \le\pi\) 
Thus, we have
cos^-1 \((-\frac{1}{\sqrt2})\) = \(\frac{3\pi}{4}\), since \(\frac{3\pi}{4}\)\(\in[0,\pi]\)cos\(\frac{3\pi}{4}\) = cos\((\pi=\frac{\pi}{4})=-cos \frac{\pi}{4}=-\frac{1}{\sqrt2}\)

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 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

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

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

\(LHS={ tan }^{ -1 }x+{ tan }^{ -1 }\frac { 2x }{ 1-{ x }^{ 2 } } \)
= \({ tan }^{ -1 }\left( \frac { x+\frac { 2x }{ 1-{ x }^{ 2 } } }{ 1-x\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x(1-{ x }^{ 2 })+2x }{ 1-{ x }^{ 2 } } }{ \frac { 1-{ x }^{ 2 }-2{ x }^{ 2 } }{ 1-{ x }^{ 2 } } } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x-{ x }^{ 3 }+2x }{ 1-{ x }^{ 2 } } }{ \frac { 1-3x^{ 2 } }{ 1-{ x }^{ 2 } } } \right) \quad \left[ \because |x|<\frac { 1 }{ \sqrt { 3 } } \right] \)
= \({ tan }^{ -1 }\left( \frac { 3x-{ x }^{ 3 } }{ 1-{ x }^{ 2 } } \times \frac { 1-{ x }^{ 2 } }{ 1-{ 3x }^{ 2 } } \right) \)
If 3x² < 1
⇒ \( |x|<\frac { 1 }{ \sqrt { 3 } }\)

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)