Question

If $f\left(x\right)={\tan }^{-1}(\sin x+\cos x),0\le x\le \pi$ Determine the intervals of monotonicity.

Answer 1

$\frac{dy}{dx}=\frac{1}{1+{(\sin x+\cos x)}^2}(\cos x-\sin x)$

solving and using trigonometric identities

$=\frac{\cos x-\sin x}{2+\sin 2x}=\frac{\sqrt{2}\cos (x+\pi /4)}{2+\sin 2x}$

The denominator $2+\sin 2x$ is always +ive.

$\therefore \frac{dy}{dx}>0$ or $<0$ when $\cos x-\sin x>0$ or $<0$ or $\sqrt{2}\cos (x+\pi /4)>0$ or $<0$ in $0\le x\le \pi .$

Hence the same answer as given by (1) and (3) only of part (c), and not by (2) as

$0\le x\le \pi .$