Question

The sum of the absolute maximum and absolute minimum values of the function $f(x)=\tan ^{-1}(\sin x-\cos x)$ in the interval $[0, \pi]$ is

• A. 0
• B. $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}$
• C. $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}$
• D. $\frac{-\pi}{12}$

Answer 1

Answer: (C)

$ f(x)=\tan ^{-1}(\sin x-\cos x)$
$ f^{\prime}(x)=\frac{\cos x+\sin x}{(\sin x-\cos x)^2+1}=0$
$ \therefore x=\frac{3 \pi}{4}$

$\text{sum}=\tan ^{-1} \sqrt{2}-\frac{\pi}{4} $
$ =\cos ^{-1} \frac{1}{\sqrt{3}}-\frac{\pi}{4}$