Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B)

${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)+{\cot }^{-1}\left(3\right)$
We have ${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)={\cot }^{-1}2$
$\therefore$ from equation (1), we have
${\cos }^{-1}2+{\cot }^{-1}3={\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$
$={\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\frac{1}{3}}\right)$
$={\tan }^{-1}\left(\frac{5/6}{\frac{6-1}{6}}\right)={\tan }^{-1}1=\frac{\pi }{4}$