The function f(x) = tan^{-1}(sin x + cos x) i | vedclass

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(D) Given $f(x) = 	an^{-1}(\sin x + \cos x)$.To find where the function is increasing,we calculate the derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin

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$(-\frac{\pi}{2}, \frac{\pi}{4})$

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(D) Given $f(x) = 	an^{-1}(\sin x + \cos x)$.To find where the function is increasing,we calculate the derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x)} = \frac{\cos x - \sin x}{2 + \sin 2x}$.For the function to be increasing,we require $f'(x) > 0$.Since $2 + \sin 2x > 0$ for all $x$,we must have $\cos x - \sin x > 0$.$\cos x > \sin x$.Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > 	an x$,which implies $x Considering the domain of the function,the interval where $f'(x) > 0$ is $(-\frac{\pi}{2}, \frac{\pi}{4})$.

The function $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in

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