prove that f(x) = tan-1(sin x+ cos x) is an increasing function in (-π/2,π/4)

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$$\begin{gathered} \mathrm{Any} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{an} \mathrm{increasing} \mathrm{function} \mathrm{in} \mathrm{given} \mathrm{interval} \left(\mathrm{a},\mathrm{b}\right) \mathrm{if} \mathrm{f}\text{'}\left(\mathrm{x}\right)>0 \mathrm{for} \\ \mathrm{all} \mathrm{x}\in \left(\mathrm{a},\mathrm{b}\right). \\ \mathrm{Now} \\ \mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1}\left(\mathrm{sinx}+\mathrm{cosx}\right) \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\tan }^{-1}\left(\mathrm{sinx}+\mathrm{cosx}\right)\right\} \\ =\frac{1}{1+{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^2}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}+\mathrm{cosx}\right) \\ =\frac{\mathrm{cosx}-\mathrm{sinx}}{1+{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^2} \\ \mathrm{Now} \mathrm{denominator} \mathrm{is} \mathrm{positive} \mathrm{for} \mathrm{all} \mathrm{values} \mathrm{of} \mathrm{x}, \mathrm{hence} \mathrm{we} \mathrm{should} \mathrm{concentrate} \mathrm{on} \mathrm{numerator}. \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\sqrt{2}\left\{\frac{\mathrm{cosx}.{\displaystyle \frac{1}{\sqrt{2}}}-\mathrm{sinx}.{\displaystyle \frac{1}{\sqrt{2}}}}{1+{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^2}\right\} \\ =\sqrt{2}\left\{\frac{\mathrm{cosx}.{\displaystyle \cos \frac{\mathrm{\pi }}{4}}-\mathrm{sinx}.{\displaystyle \sin \frac{\mathrm{\pi }}{4}}}{1+{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^2}\right\} \\ =\sqrt{2}\left\{\frac{\cos \left(\mathrm{x}+{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{1+{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^2}\right\} \left[\mathrm{As} \mathrm{cosA}.\mathrm{cosB}-\mathrm{sinA}.\mathrm{sinB}=\cos \left(\mathrm{A}+\mathrm{B}\right)\right] \\ \mathrm{Now} \mathrm{given} \mathrm{that} \\ -\frac{\mathrm{\pi }}{2}<\mathrm{x}<\frac{\mathrm{\pi }}{4} \\ \Rightarrow -\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}<\mathrm{x}+\frac{\mathrm{\pi }}{4}<\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\mathrm{x}+\frac{\mathrm{\pi }}{4}<\frac{\mathrm{\pi }}{2} \\ \mathrm{Now} \mathrm{cos\alpha } \mathrm{is} \mathrm{positive} \mathrm{when} \mathrm{\alpha }\in \left(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right) \left[\mathrm{Here} \mathrm{\alpha }=\mathrm{x}+\frac{\mathrm{\pi }}{4}\right] \\ \mathrm{Hence} \mathrm{f}\text{'}\left(\mathrm{x}\right)>0 \mathrm{for} \mathrm{all} \mathrm{x}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{an} \mathrm{increasing} \mathrm{function} \mathrm{for} \mathrm{x}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{4}\right). \end{gathered}$$

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