# Question #58780

Nov 1, 2017

$R H S = \frac{{\sin}^{2} x}{2 + \sin 2 x \csc x}$

$= {\left(2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)\right)}^{2} / \left(2 + 2 \sin x \cos x \times \frac{1}{\sin} x\right)$

$= \frac{4 {\sin}^{2} \left(\frac{x}{2}\right) {\cos}^{2} \left(\frac{x}{2}\right)}{2 \left(1 + \cos x\right)}$

$= \frac{4 {\sin}^{2} \left(\frac{x}{2}\right) {\cos}^{2} \left(\frac{x}{2}\right)}{2 \cdot 2 {\cos}^{2} \left(\frac{x}{2}\right)}$

$= {\sin}^{2} \left(\frac{x}{2}\right) = L H S$

Nov 1, 2017

Formulas to be used in the proof.

1. $\sin 2 x = 2 \sin x \cos x$

2. $\sin \left(\frac{1}{2} x\right) = \pm \sqrt{\frac{1 - \cos x}{2}}$
$\therefore {\sin}^{2} \left(\frac{1}{2} x\right) = \frac{1 - \cos x}{2}$

3. $\csc x = \frac{1}{\sin} x$
4. ${\sin}^{2} x + {\cos}^{2} x = 1$
${\sin}^{2} x = 1 - {\cos}^{2} x$

Right Hand Side:

${\sin}^{2} \frac{x}{2 + \sin 2 x \csc x} = {\sin}^{2} \frac{x}{2 + 2 \sin x \cos x \csc x}$

$= {\sin}^{2} \frac{x}{2 + 2 \cancel{\sin} x \cos x \cdot \frac{1}{\cancel{\sin}} x}$

$= \frac{1 - {\cos}^{2} x}{2 + 2 \cos x}$

$= \frac{\left(1 - \cos x\right) \left(1 + \cos x\right)}{2 \left(1 + \cos x\right)}$

$= \frac{\left(1 - \cos x\right) \cancel{1 + \cos x}}{2 \cancel{1 + \cos x}}$

$= \frac{1 - \cos x}{2}$

$= {\sin}^{2} \left(\frac{1}{2}\right) x$

$\therefore =$Left Hand Side