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Show that f (x) = tan^{-1}(\sin x + \cos x)is an increasing function in \left(0,\frac{\pi}{4}\right)

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Given: f (x) = tan^{-1}(\sin x + \cos x)

To show: the given function is increasing in  \left(0,\frac{\pi}{4}\right)

Explanation: Given

f (x) = tan^{-1}(\sin x + \cos x)

First derivative is applied with respect to x,

\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\tan ^{-1}(\sin \mathrm{x}+\cos \mathrm{x})\right)}{\mathrm{dx}}
Using the differentiation rule for tan ^{-1}, results into
\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{(\sin \mathrm{x}+\cos \mathrm{x})^{2}+1} \cdot \frac{\mathrm{d}(\sin \mathrm{x}+\cos \mathrm{x})}{\mathrm{dx}}
Now use the sum rule of differentiation,

\begin{aligned} &f^{\prime}(x)=\frac{1}{(\sin x+\cos x)^{2}+1}\left[\frac{d(\sin x)}{d x}+\frac{d(\cos x)}{d x}\right]\\ &\text { But the derivative of } \sin \mathrm{X}=\cos \mathrm{x} \text { and that } \text { of } \cos \mathrm{x}=-\sin \mathrm{x}, \mathrm{s} \circ\\ &\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{(\sin \mathrm{x}+\cos \mathrm{x})^{2}+1}[\cos \mathrm{x}+(-\sin \mathrm{x})]\\ &\text { Expanding }(\sin x+\cos x)^{2}, \text { we get }\\ &f^{\prime}(x)=\frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+1}\\ &\text { But } \sin ^{2} x+\cos ^{2} x=1 \text { and } 2 \sin x \cos x=\sin 2 x, \text { thus the equation given above gets converted }\\ &\text { into, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos \mathrm{x}-\sin \mathrm{x}}{1+\sin 2 \mathrm{x}+1}\\ &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos \mathrm{x}-\sin \mathrm{x}}{\sin 2 \mathrm{x}+2} \end{aligned}

To make f(x) to be increasing function,
\\f^{\prime}(x) \geq 0$ \\$\Rightarrow \frac{\cos x-\sin x}{\sin 2 x+2} \geq 0$ \\$\Rightarrow \cos x-\sin x \geq 0$ \\$\Rightarrow \cos x \geq \sin x
But this is possible only when  \mathrm{x} \in\left(0, \frac{\pi}{4}\right)
Hence, the given function is increasing function in  \left(0, \frac{\pi}{4}\right)

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