Mathematics | VITEEE 2016 | Discussion Page

The value of $\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cot^{-1}(3)$ is

$\frac{\pi}{6}$

$\frac{\pi}{4}$

$\frac{\pi}{3}$

$\frac{\pi}{2}$

Option B

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

$\cos^{-1}2+\cot^{-1}3=\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}$

$\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\right)$

= $\tan^{-1}\left(\frac{\frac{5/6}{6-1}}{6}\right)=\tan^{-1}1=\frac{\pi}{4}$

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