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

We know that \(sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) =cos^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
Thus, \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =\frac { 1 }{ \sqrt { 1+x^{ 2 } } } \) ...(1)
Let \(\cot ^{-1}\left(\frac{3}{4}\right)=\theta\). Then \(\cot \theta=\frac{3}{4}\) and so \(\theta\) is cute. 
From the diagram, we get,
Hence \(\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}=\sin \theta=\frac{4}{5}\) ................ (2)
Using (1) and (2) in the given equation, we \(\frac { 1 }{ \sqrt { 1+x^{ 2 } } } =\frac { 4 }{ 5 } \) \(\sqrt{1+x^2}=\frac{5}{4}\)
Thus, x = \(\pm\frac{3}{4}\)

tan^-1 \(\left( \frac { 1-x }{ 1+x } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\)x gives tan^-1 1-tan^-1 x = \(\frac{1}{2}\)tan^-1x. 
Therefore, \(\frac{\pi}{4}=\frac{3}{2}tan^{-1}\)x, which in turn reduces to tan⁻¹ = \(\frac{\pi}{6}\)
Thus, x = tan\(\frac{\pi}{6}=\frac{1}{\sqrt3}\)

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

 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]

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

Given 
\({ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) ={ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore x=\frac { 1 }{ \sqrt { 3 } } \)

Let sec^-1\(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\theta \).
Then, sec\(\theta\) = \(-\frac{2}{\sqrt3}\) where \(\theta\in[0,\pi]\)\{\(\frac{\pi}{2}\)}.
Thus, cos \(\theta =-\frac{\sqrt{3}}{2}\).
Now, \(cos\frac { 5\pi }{ 6 } =cos\left( \pi -\frac { \pi }{ 6 } \right) =-cos\left( \frac { \pi }{ 6 } \right) =-\frac { \sqrt { 3 } }{ 2 } .\) 
Hence, Sec-¹ \(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\frac { 5\pi }{ 6 } \)

\({ tan }^{ -1 }\sqrt { x(x+1) } +{ sin }^{ -1 }\sqrt { { x }^{ 2 }+x+1 } =\frac { \pi }{ 2 } \)
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.

LHS = \({ tan }^{ -1 }\left( \frac { m }{ n } \right) -{ tan }^{ -1 }\left( \frac { m-n }{ m+n } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { m }{ n } -\frac { m-n }{ m+n } }{ 1+\frac { m }{ n } \left( \frac { m-n }{ m+n } \right) } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { m(m+n)-n(m-n) }{ m(m+n) } }{ \frac { n(m+n)+m(m-n) }{ n(m+n) } } \right) \)

= \({ tan }^{ -1 }\left( \frac { { m }^{ 2 }+{ n }^{ 2 } }{ { m }^{ 2 }+{ n }^{ 2 } } \right) ={ tan }^{ -1 }(1)\)
= \(\frac { \pi }{ 4 } =RHS\)
Hence proved.

\(tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) \)
Let \(sin^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\)

\(\Rightarrow \frac { 3 }{ 5 } =sinx\)
\(\therefore tanx=\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\({ cot }^{ -1 }\left( \frac { 3 }{ 2 } \right) =y\)
\(\Rightarrow \frac { 3 }{ 2 } =coty\Rightarrow tany=\frac { 2 }{ 3 } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =tan(x+y)\)
\(\frac { tanx+tany }{ 1-tanxtany } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =\frac { 17 }{ 6 } \)

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

= tan^-1(1)
= \(\frac { \pi }{ 4 } \)

By definition, cot^-1x\(\in(0,\pi)\)
Therefore, cot^-1\((\frac{1}{7})\) = \(\theta\) implies \(\theta \in(0,\pi)\)
But cot^-1\((\frac{1}{7})\) = \(\theta\) implies cot \(\theta\) = \(\frac{1}{7}\) and hence tan \(\theta\) = 7 and \(\theta\) is acute.
Using tan \(\theta\) = \(\frac{7}{1}\), We construct a right triangle as shown .
Then, we have, cos \(\theta\) = \(\frac{1}{5\sqrt2}\).

Periodic function