Answer:
Option B
Explanation:
Consider
$\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cot^{-1}3$ ..........(i)
We have, $\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)=\cot^{-1}2$
$\therefore$ From equation (i) , we have
$\cos^{-1}2+\cot^{-1}3=\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}$
$\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\right)$
2
= $\tan^{-1}\left(\frac{\frac{5/6}{6-1}}{6}\right)=\tan^{-1}1=\frac{\pi}{4}$
