JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
The value of $\tan^{-1}\left(\frac{\cos\left(\frac{15\pi}{4}\right)-1}{\sin\left(\frac{\pi}{4}\right)}\right)$ is equal to :
Inverse Trigonometric Functions
Solution
$\cos\frac{15\pi}{4}=\cos\frac\pi4=\frac{1}{\sqrt2}$, $\sin\frac\pi4=\frac1{\sqrt2}$. Argument $=\frac{1/\sqrt2-1}{1/\sqrt2}=1-\sqrt2=-\tan\frac\pi8$. $\tan^{-1}(-\tan\frac\pi8)=-\frac\pi8$. Option (B).