prove that $ 2 \arctan({\csc \arctan x - \tan \text{arccot }x}) = \arctan x $

Question

Prove that $ 2 \arctan({\csc (\arctan x) -\tan (\text{arccot }x)}) = \arctan x $

x is not equal to zero.

So, to solve this I tried I made two condition

$ x \gt 0 $ and $ x \lt 0 $

If $ x \gt 0 $

$= 2 \arctan({\csc(\arctan x) - \tan(\text{arccot } x)}) $

$ = 2 \arctan\left(\dfrac{\sqrt{1+x^2}}{x} - \dfrac{1}{x}\right) $

putting $ x = \tan\theta $

$ = 2 \arctan x $

if $ x \lt 0 $

putting $ x = -|x |$

$= 2 \arctan(\csc(\arctan (- x)) - \tan (\text{arccot }( - x)) $

$ = 2 \arctan\left( - {\dfrac{\sqrt{1+x^2}}{x} + \dfrac{1}{x} }\right) $

putting $ x = \tan \theta $

$ = 2 \arctan\left( {\dfrac{1-\sqrt{1+\tan ^2\theta }}{\tan \theta} }\right) $

$ =- 2 \arctan\left( { \dfrac{1-\cos \theta }{\tan \theta \cos \theta } }\right) $

$ =- 2 \arctan x $

What is wrong with it?

Answer 1

Hint:

Let $\arctan x=2y,-\pi/2<2y<\pi/2,y\ne0$

$\tan($arccot $x)=\tan(\pi/2-2y)=\cot2y$

$\csc2y-\cot2y=\cdots=\tan y$

Answer 2

Let $\arctan x=y$

$\tan y=x,-\pi/2<y<\pi/2$

$\tan($arccot $x)=\tan(\pi/2-y)$ $=\cot y=\dfrac1x$

$\sec y=+\sqrt{1+x^2}$

$\csc\arctan x)=\csc y=\dfrac{\sec y}{\tan y}=?$