# How do you verify sin^2 (-x) / tan^2 x = cos^2 x?

Apr 16, 2016

We will be using the following:

• $\sin \left(- x\right) = - \sin \left(x\right)$

• $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

Then, when $\cos \left(x\right) \ne 0$ and $\sin \left(x\right) \ne 0$, we have

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

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

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

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

$= {\cos}^{2} \left(x\right)$