Show that f(x)= ^ -1 ( + ) is an increasing function in (0, 4 ).

Question

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

✓

Answer

We have, $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$
$\therefore\ \text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}\cdot(\cos\text{x}-\sin\text{x})$
$=\frac{1}{1+\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}.\cos\text{x}}(\cos\text{x}-\sin\text{x})$
$=\frac{1}{(2+\sin2\text{x})}(\cos\text{x}-\sin\text{x})$
$\big[\because\sin2\text{x}=2\sin\text{x}\cos\text{x and }\sin^2\text{x}+\cos^2\text{x}=1\big]$
For $\text{f}'(\text{x})\geq0.$
$\frac{1}{(2+\sin2\text{x})}\cdot(\cos\text{x}-\sin\text{x})\geq0$
$\Rightarrow\ \cos\text{x}-\sin\text{x}\geq0$ $\Big[\because(2+\sin2\text{x})\geq0\text{ in }\Big(0,\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\ \cos\text{x}\geq\sin\text{x}$
Which is true, if $\text{x}\in\Big(0,\frac{\pi}{4}\Big)$
Hence, f(x) is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

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