JEE Main 2021 (Online) 27th August Evening ShiftQuestion 1
Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x)=\tan^{-1}(\sin x+\cos x)$ in $\left[0,\frac{\pi}{2}\right]$. Then the value of $\tan(M-m)$ is equal to :
Inverse Trigonometric Functions
Solution
$\sin x+\cos x=\sqrt2\sin(x+\frac\pi4)$ on $[0,\frac\pi2]$ ranges over $[1,\sqrt2]$. $M=\tan^{-1}\sqrt2$, $m=\tan^{-1}1=\frac\pi4$. $\tan(M-m)=\frac{\sqrt2-1}{1+\sqrt2}=(\sqrt2-1)^2=3-2\sqrt2$. Option (C).