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The function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function in
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Verified Answer
The correct answer is:
$\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$
Since, $f(x)=\tan ^{-1}(\sin x+\cos x)$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+(\sin \mathrm{x}+\cos \mathrm{x})^2}(\cos \mathrm{x}-\sin \mathrm{x})$
$=\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}$
$f(x)$ is increasing if $f^{\prime}(x)>0 \Rightarrow \cos \left(x+\frac{\pi}{4}\right)>0$
$\Rightarrow \quad-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2} \Rightarrow-\frac{3 \pi}{2} < x+\frac{\pi}{4}$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+(\sin \mathrm{x}+\cos \mathrm{x})^2}(\cos \mathrm{x}-\sin \mathrm{x})$
$=\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}$
$f(x)$ is increasing if $f^{\prime}(x)>0 \Rightarrow \cos \left(x+\frac{\pi}{4}\right)>0$
$\Rightarrow \quad-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2} \Rightarrow-\frac{3 \pi}{2} < x+\frac{\pi}{4}$
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