\(tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 2a }{ 1+{ a }^{ 2 } } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ a }^{ 2 } }{ 1+{ a }^{ 2 } } \right) \right] \)
Let a = tan 0
Now, \(tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 2a }{ 1+{ a }^{ 2 } } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ a }^{ 2 } }{ 1+{ a }^{ 2 } } \right) \right] =tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right) \right] \)
\(tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }(sin2\theta )+\frac { 1 }{ 2 } { cos }^{ 1 }(cos2\theta ) \right] =tan[2\theta ]=\frac { 2tan\theta }{ 1-{ tan }^{ 2 }\theta } =\frac { 2a }{ 1-{ a }^{ 2 } } \)

Consider \({ tan }^{ -1 }\left( x-1 \right) +{ tan }^{ -1 }\left( x+1 \right) \)
= \({ tan }^{ -1 }\left( \frac { x-1+x+1 }{ 1-(x-1)(x+1) } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-\left( { x }^{ 2 }-1 \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { x-1+x+1 }{ 1-(x-1)(x+1) } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-\left( { x }^{ 2 }-1 \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 }+1 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) \)
\(\therefore { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) +{ tan }^{ -1 }\left( x \right) ={ tan }^{ -1 }\left( 3x \right) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( 3x \right) -{ tan }^{ -1 }\left( x \right) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( \frac { 3xx }{ 1+3x^{ 2 } } \right) \)
\(\Rightarrow \frac { 2x }{ 2-{ x }^{ 2 } } =\frac { 2x }{ 1+{ 3x }^{ 2 } } \)
x(1+ 3x²) = x(2-x²)
x[1+ 3x²-2 + x²] = 0
x(4x²-1) = 0
x(2x+ 1) (2x-1) = 0
x = 0, x = \(\frac{1}{2}\), x = -\(\frac{1}{2}\) are the roots 
Hence, there are 3 solutions

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

\({ tan }^{ -1 }x+{ tan }^{ -1 }y+{ tan }^{ -1 }z={ tan }^{ -1 }\left( \frac { x+y+z-xyz }{ 1-xy-yz-zx } \right) \)
Given \({ tan }^{ -1 }x+{ tan }^{ -1 }x+{ tan }^{ -o }y+{ tan }^{ -1 }z=\pi \)
\(\Rightarrow \pi ={ tan }^{ -1 }\left( \frac { x+y+zxyz }{ 1-xy-yz-zx } \right) \)
\(\Rightarrow tan\pi =\frac { x+y+z-xyz }{ 1-xy-yz-zx } \)
\(\Rightarrow 0=\frac { x+y+z-xyz }{ 1-xy-yz-zx } \quad [\therefore tan\pi =0]\)
\(\Rightarrow x+y+z-xyz=0\)
\(\Rightarrow x+y+z=xyz\)

We know that \({ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \)
\(\therefore LHS={ tan }^{ -1 }(x)+{ tan }^{ -1 }(y)+{ tan }^{ -1 }(z)\)
= \({ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) +{ tan }^{ -1 }\left( z \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y }{ 1-xy } +z }{ 1-z\left( \frac { x+y }{ 1-xy } \right) } \right) \) by(1)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y+z(1-xy) }{ 1-xy } }{ \frac { (1-xy)-z(x+y) }{ 1-xy } } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y+z-xyz }{ 1-xy } }{ \frac { 1-xy-zx-zy }{ 1-xy } } \right) \)

= \({ tan }^{ -1 }\left( \frac { x+y+z-xyz }{ 1-xy-yz-zx } \right) =RHS\)
Hence proved.

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

 f(x) = sin^-1 \((\frac{|x|-2}{3})+cos^-1(\frac{1-|x|}{4})\)
From the definition of sin^-1
\(-1\le \frac { \left| x \right| -2 }{ 3 } \le 1\)
\(\Rightarrow -3\le \left| x \right| -2\le 3\)
\(\Rightarrow -3+2\le \left| x \right| \le \left| x \right| \le 3+2\)
\(\Rightarrow -1\le \left| x \right| \le 5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(\Rightarrow 0\le \left| x \right| and\left| x \right| \le 5\)
\(\Rightarrow \left| x \right| \ge 0and-5\le x\le 5\)
From the definition of cos^-1x.
\(-1\le \frac { 1-\left| x \right| }{ 4 } \le 1\)
\(\Rightarrow -4\le 1-\left| x \right| \le 4\)
\(\Rightarrow -4-1\le \left| x \right| \le 4-1\)
\(\Rightarrow -5\le -\left| x \right| \le 3\)
\(\Rightarrow -3\le \left| x \right| >5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(-5\le |x|\le 5\)
From (1) & (2),
Domain is [-5, 5]

\({ cot }^{ -1 }(1)+{ sin }^{ -1 }\left( -\frac { \sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }(-\sqrt { 2 } )\)
Let \(\Rightarrow cot\ x=1\Rightarrow tan\ x=1\Rightarrow tan\ x=tan\frac { \pi }{ 4 } \)
\(\Rightarrow x=\frac { \pi }{ 4 } \)
\({ sin }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } \right) =y\Rightarrow sin\ y=\frac { -\sqrt { 3 } }{ 2 } =-sin\frac { \pi }{ 3 } \)
\(\Rightarrow sin\ y=sin\left( { \frac { -\pi }{ 3 } } \right) \)
\(\Rightarrow y=\frac { -\pi }{ 3 } \)
\({ sec }^{ -1 }\left( -\sqrt { 2 } \right) =z\)
\(\Rightarrow sec\ z=-\sqrt { 2 } \Rightarrow cosz=\frac { -1 }{ \sqrt { 2 } } \)
\(\Rightarrow cos\ z=-cos\frac { \pi }{ 4 } \)
\(\Rightarrow cos\ z=cos\left( \pi -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow cos\ z=cos\left( \frac { 3\pi }{ 4 } \right) \)
\(\Rightarrow z=\frac { 3\pi }{ 4 } \)
\(\therefore { cot }^{ -1 }(1)+{ sin }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }\left( -\sqrt { 2 } \right) \)
= \(\frac { \pi }{ 4 } -\frac { \pi }{ 3 } -\frac { 3\pi }{ 4 } \)

= \(-\frac { 5\pi }{ 6 } \)
\(\therefore\) \({ cot }^{ -1 }(1)+{ sin }^{ -1 }\left( -\frac { \sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }(-\sqrt { 2 } )\) =\(-\frac { 5\pi }{ 6 } \)

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

= tan^-1(1)
= \(\frac { \pi }{ 4 } \)

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\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

\({ sin }^{ -1 }(-1)+{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }(2)\)
Let \({ sin }^{ -1 }\left( -1 \right) =x\)
\(\Rightarrow -1=sin\ x\)
\(\Rightarrow sin\ x=-1=-sin\frac { \pi }{ 2 } =sin\left( \frac { -\pi }{ 2 } \right) \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\Rightarrow \frac { 1 }{ 2 } =cos\ y\Rightarrow cos\frac { \pi }{ 3 } \)
\(\Rightarrow y=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ -1 }\left( -1 \right) +{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }\left( 2 \right) \)
\(\frac { -\pi }{ 2 } +\frac { \pi }{ 3 } +{ cot }^{ -1 }\left( 2 \right) \)
\({ cot }^{ -1 }(2)+\frac { -3\pi +2\pi }{ 0 } ={ cot }^{ -1 }\left( 2 \right) -\frac { \pi }{ 6 } \)