Prove that $y = \cfrac{{4\sin \theta }}{{\left( {2 + \cos \theta } \right)}} - \theta$ is an increasing function of $\theta$ in $\left[ {0,\cfrac{\pi }{2}} \right]$.

We have, $y = \cfrac{{4\sin \theta }}{{2 + \cos \theta }} - \theta ,\theta \in \left[ {0,\cfrac{\pi }{2}} \right]$
$\Rightarrow \cfrac{{dy}}{{d\theta }} = 4\left\{ {\cfrac{{(2 + \cos \theta )\cos \theta - \sin \theta ( - \sin \theta )}}{{{{(2 + \cos \theta )}^2}}}} \right\} - 1$

$= \cfrac{{4(2\cos \theta + 1)}}{{{{(2 + \cos \theta )}^2}}} - 1 = \cfrac{{8\cos \theta + 4 - {{(2 + \cos \theta )}^2}}}{{{{(2 + \cos \theta )}^2}}}$
$= \cfrac{{4\cos \theta - {{\cos }^2}\theta }}{{{{(2 + \cos \theta )}^2}}} = \cfrac{{\cos \theta (4 - \cos \theta )}}{{{{(2 + \cos \theta )}^2}}}$

Now $\cos \theta > 0\,\,{\rm{in}}\,\,\left[ {0,\cfrac{\pi }{2}} \right]$ ; $4 - \cos \theta > 0\,\,{\rm{in}}\,\,\left[ {0,\cfrac{\pi }{2}} \right]$

and ${(2 + \cos \theta )^2} > 0$ in $\left[ {0,\cfrac{\pi }{2}} \right]$ [Being a perfect square]

Therefore, $\cfrac{{dy}}{{d\theta }} > 0$ for all $\theta \in \left[ {0,\cfrac{\pi }{2}} \right]$

Hence we can say that $y$ is strictly increasing function in $\left[ {0,\cfrac{\pi }{2}} \right]$