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

This question has appeared in model paper 2012

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=\large\frac{4\sin \theta}{(2+\cos \theta)}$$-\theta Differentiating w.r.t \theta we get, f'(x)=\large\frac{(2+\cos \theta).4\cos \theta-4\sin\theta(0-\sin\theta)}{(2+\cos \theta)^2}$$-1$
$\qquad=\large\frac{8\cos \theta+4\cos^2\theta+4\sin^2\theta}{(2+\cos \theta)^2}$$-1 (But \sin^2\theta+\cos^2\theta=1) \qquad=\large\frac{8\cos \theta+4}{(2+\cos \theta)^2}$$-1$
Step 2:
On simplifying we get,
$f'(x)=\large\frac{8\cos \theta+4-(2+\cos \theta)^2}{(2+\cos\theta)^2}$
$\qquad=\large\frac{4\cos \theta-cos^2\theta}{(2+\cos\theta)^2}$
$\qquad=\large\frac{\cos \theta(4-cos^2\theta)}{(2+\cos\theta)^2}$
$\Rightarrow \cos \theta(4-\cos ^2\theta)>0$
$(4-\cos^2\theta) >0$ or $\cos^2\theta$ is not $>1$
$\Rightarrow \theta \in (0,\large\frac{\pi}{2})$