Question

Identify the pair(s) of functions which are identical.
i) $y=\sin \left(arc\tan x\right)$; $y=\frac{x}{\sqrt{1+x^2}}$
(ii) $y=\cos \left(arc\tan x\right)$; $y=\sin \left(arc\cot x\right)$.
(iii) $y=\tan \left({\cos }^{-1}x\right)$; $y=\frac{\sqrt{1-x^2}}{x}$
iv) $y=\tan \left({\cot }^{-1}x\right)$; $y=\frac{1}{x}$

Answer 1

$i)$ $\sin \left(arc\tan x\right)=\sin \left({\tan }^{-1}x\right)=\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)=\frac{x}{\sqrt{1+x^2}}$.

$ii)$ $\cos \left(arc\tan x\right)$

$=\cos \left(ta{n}^{-1}x\right)$

$=\cos (\frac{\pi }{2}-{\cot }^{-1}x)$

$=\sin \left(arc\cot x\right)$

$iii\left)\tan \right({\cos }^{-1}x)$

$=\tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)$

$=\frac{\sqrt{1-x^2}}{x}$

$iv\left)\tan \right({\cot }^{-1}x)=\tan \left({\tan }^{-1}\frac{1}{x}\right)=\frac{1}{x}$.

So, given all the pairs are identical.