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

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(C) Given the function $f(x) = 	an^{-1}(\sin x + \cos x)$.Differentiating with respect to $x$,we get:$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot

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$(0, \frac{\pi}{4})$

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(C) Given the function $f(x) = 	an^{-1}(\sin x + \cos x)$.Differentiating with respect to $x$,we get:$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.For the function to be increasing,we require $f'(x) > 0$.Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,the sign of $f'(x)$ depends on the numerator $(\cos x - \sin x)$.We need $\cos x - \sin x > 0$,which implies $\cos x > \sin x$.Dividing by $\cos x$ (assuming $\cos x > 0$ in the first quadrant),we get $1 > 	an x$,or $	an x This holds true for $x \in (0, \frac{\pi}{4})$.Thus,the function is strictly increasing in the interval $(0, \frac{\pi}{4})$.

The function $f(x) = \tan^{-1}(\sin x + \cos x), x > 0$ is an increasing function in the interval.....

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