Prove that is an increasing function of θ in.

Solution:

We have,

$y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$

$\therefore \frac{d y}{d x}=\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1$

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

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

Now, $\frac{d y}{d x}=0$

$\Rightarrow \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}=1$

$\Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta$

$\Rightarrow \cos ^{2} \theta-4 \cos \theta=0$

$\Rightarrow \cos \theta(\cos \theta-4)=0$

$\Rightarrow \cos \theta=0$ or $\cos \theta=4$

Since $\cos \theta \neq 4, \cos \theta=0$

$\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}$

Now,

$\frac{d y}{d x}=\frac{8 \cos \theta+4-\left(4+\cos ^{2} \theta+4 \cos \theta\right)}{(2+\cos \theta)^{2}}=\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}$

$\therefore \cos \theta(4-\cos \theta)>0$ and also $(2+\cos \theta)^{2}>0$

$\begin{aligned} \text { In interval }\left(0, \frac{\pi}{2}\right), \text { we have } \cos \theta>0 . \text { Also, } 4>\cos \theta \Rightarrow 4-\cos \theta>0 . & \Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0 \\ & \Rightarrow \frac{d y}{d x}>0 \end{aligned}$

Therefore, $y$ is strictly increasing in interval $\left(0, \frac{\pi}{2}\right)$.

Also, the given function is continuous at $x=0$ and $x=\frac{\pi}{2}$.

Hence, $y$ is increasing in interval $\left[0, \frac{\pi}{2}\right]$.