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(C) To find the interval where $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\

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$\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$

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(C) To find the interval where $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x)$.$f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}$.For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,we only need the numerator to be positive:$\cos x - \sin x > 0$.$\cos x > \sin x$.Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > 	an x$,which implies $	an x This inequality holds when $x \in \left(-\frac{\pi}{2}, \frac{\pi}{4}ight)$.Comparing this with the given options,the interval $\left(-\frac{3\pi}{4}, \frac{\pi}{4}ight)$ contains the region where the function is increasing,and specifically,the function increases within the range provided in option $C$.

The interval in which the function $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is an increasing function,is

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