# Can anyone help prove this trigonometric identity?

## $\sin \frac{x}{1 + \cos x} + \cos \frac{x}{\sin} x = \csc x$

Nov 29, 2016

Cross multiply and simplify LHS (as below)

#### Explanation:

To prove: $\sin \frac{x}{1 + \cos x} + \cos \frac{x}{\sin} x = \csc x$

$L H S = \sin \frac{x}{1 + \cos x} + \cos \frac{x}{\sin} x$

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

Since: ${\sin}^{2} x + {\cos}^{2} x = 1$

$L H S = \frac{1 + \cos x}{\sin x \left(1 + \cos x\right)}$

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

$= \frac{1}{\sin} x = \csc x = R H S$

Nov 29, 2016

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

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

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

$= \frac{1 - \cos x}{\sin} x + \cos \frac{x}{\sin} x$

$= \frac{1 - \cos x + \cos x}{\sin} x$

$= \frac{1}{\sin} x$

$= \csc x$