Question
Prove that y = $\frac{\text{4 sin}\theta}{\text{2 + cos}\theta}-\theta$ is an increasing function in $\Bigg[0,\frac{\pi}{2}\Bigg].$

Answer

$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{(2 + cos}\theta)4\cos\theta-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^{2}}$
$=\frac{8\cos\theta+4(\cos^{2}\theta+\sin^{2}\theta)-(4+\cos^{2}\theta+4\cos\theta)}{(2+\cos\theta)^{2}}$
$=\frac{4\cos\theta-cos^{2}\theta}{(2+\cos\theta)^{2}}=\frac{4-\cos\theta}{(2+\cos\theta)^{2}}\cdot\cos\theta$
since, $\frac{4-\cos\theta}{(2+cos\theta)^{2}}>0 \text{ for }\forall\theta\text{ and cos}\theta\geq0\text{ in }\Bigg[0,\frac{\pi}{2}\Bigg]$
$\therefore\frac{\text{dy}}{\text{d}\theta}\geq\text{ 0 in }\Bigg[0,\frac{\pi}{2}\Bigg],$ Hence the function is increasing in $\Bigg[0,\frac{\pi}{2}\Bigg]$.

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