Q. The value of $\sin^{-1} \left(\frac{1}{\sqrt{5}}\right) + \cot^{-1} \left(3\right) $ is :
Inverse Trigonometric Functions
Solution:
$\sin^{-1} \left(\frac{1}{\sqrt{5}}\right) + \cot^{-1} \left(3\right)$
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} $
