# How do you simplify (sinx/(1-cosx))+((1-cosx)/sinx)?

Mar 1, 2016

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

#### Explanation:

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

Multiply the first term by $\sin \frac{x}{\sin} x$ and the second term by $\frac{1 - \cos x}{1 - \cos x}$

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

Group terms with common denominators

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

Expand ${\left(1 - \cos x\right)}^{2}$

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

Apply the identity ${\sin}^{2} x + {\cos}^{2} x = 1$

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

Factor out $2$ from the numerator

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

Cancel common terms from the numerator and denominator

$= \frac{2}{\sin} x$

Apply the definition of the cosecant function ($\csc x = \frac{1}{\sin} x$)

$= 2 \csc x$

Mar 1, 2016

$\frac{2}{\sin} x$

#### Explanation:

Write with a common denominator

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

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

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

[using the identity : sin^2x + cos^2x = 1]

then becomes :$\frac{\left(1 + 1 - 2 \cos x\right)}{\sin x \left(1 - \cos x\right)}$

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

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