# How do you prove (cosx+ 1)/ sin^3x = cscx/(1-cosx)?

Dec 20, 2015

Let's manipulate only the right hand side in order to make it appear like the left hand side.

First, multiply by the conjugate of the denominator.

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

Note that since ${\sin}^{2} x + {\cos}^{2} x = 1$, it's true that ${\sin}^{2} x = 1 - {\cos}^{2} x$.

Thus, the expression can be rewritten as:

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

Recall that $\csc x = \frac{1}{\sin} x$.

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

Multiply the numerator and denominator by $\sin x$.

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

$= \frac{\cos x + 1}{\sin} ^ 3 x$

Voilà! This is the left hand side.

Feb 1, 2017

This is method without conjugates. It involves manipulating only the left-hand side of the equation.

$\frac{\cos x + 1}{\sin} ^ 3 x$

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

Through the Pythagorean Identity:

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

Factoring:

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

Cancelling $\frac{\cos x + 1}{1 + \cos x} = \frac{\cos x + 1}{\cos x + 1} = 1$:

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

Rewriting $\sin x$ as $\frac{1}{\csc} x$:

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

Inverting:

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

Which is the right-hand side. Thus the equality is proven.