Question

how to convert tan inverse to cos inverse formula

Answer 1

Consider the right triangle in attachment.

Applying trig ratios we get :
$\tan(t)~=~x\implies t=\tan^{-1}(x)$
$\cos(t)~=~\frac{1}{\sqrt{1+x^2}}\implies t=\cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$
$\sin(t)~=~\frac{x}{\sqrt{1+x^2}}\implies t=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$

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