Question

Prove that
$\cos \left({\tan }^{-1}\left(\sin \left({\cot }^{-1}x\right)\right)\right)=\sqrt{\frac{x^2+1}{x^2+2}}$

Answer 1

Let $co{t}^{-1}x=y$

Ref. image

then $coty=x$ and $\sin y=\frac{1}{\sqrt{1+x^2}}$

Now $y={\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

$\Rightarrow$ so $\cos \left({\tan }^{-1}\right(\sin \left({\cot }^{-1}x\right)\left)\right)=\cos \left({\tan }^{-1}\left(\sin {\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right)\right)$

$\Rightarrow$ we know that $\sin \left({\sin }^{-1}p\right)=p$

using this we get

$\Rightarrow cos\left({\tan }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right)\Rightarrow$ now ${\tan }^{-1}\frac{1}{\sqrt{1+x^2}}=p$

if $\tan p=\frac{1}{\sqrt{1+x^2}}$

Ref. image

then $\cos p=\frac{\sqrt{1+x^2}}{2+x^2}$

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

using this we get

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

$\Rightarrow \frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}=RHS$ Ans

Hence LHS = RHS.