# How do you verify the identity sqrt((1-costheta)/(1+costheta))=(1-costheta)/abssintheta?

Jan 19, 2017

All you really have to do is multiply by $\frac{\sqrt{1 - \cos \theta}}{\sqrt{1 - \cos \theta}}$:

$\implies \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \frac{\sqrt{1 - \cos \theta}}{\sqrt{1 - \cos \theta}}$

$= \frac{1 - \cos \theta}{\sqrt{\left(1 + \cos \theta\right) \left(1 - \cos \theta\right)}}$

From the difference of two squares, we get:

$= \frac{1 - \cos \theta}{\sqrt{1 - {\cos}^{2} \theta}}$

From the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, we get:

$= \frac{1 - \cos \theta}{\sqrt{{\sin}^{2} \theta}}$

From the definition $\sqrt{{u}^{2}} = | u |$, we get:

$= \textcolor{b l u e}{\frac{1 - \cos \theta}{|} \sin \theta |}$ color(blue)(sqrt"")

since the square root of ${u}^{2}$ is necessarily positive.