# How can (sin^2(-x))/(tan^2(x)) equal to cos^2x?

Aug 9, 2018

#### Explanation:

Here ,

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

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

Note that ,

$\sin \frac{- x}{\tan} x = \frac{- \sin x}{\sin \frac{x}{\cos} x} = - \sin \frac{x}{\sin} x \times \cos x = - \cos x$

Aug 9, 2018

#### Explanation:

Recall that, $\sin \left(- x\right) = - \sin x$.

$\therefore {\sin}^{2} \left(- x\right) = {\left\{\sin \left(- x\right)\right\}}^{2} = {\left(- \sin x\right)}^{2} = {\sin}^{2} x$.

$\text{Hence, the Expression} = {\sin}^{2} \frac{- x}{\tan} ^ 2 x$,

$= {\sin}^{2} x \div {\tan}^{2} x$,

$= {\sin}^{2} x \div {\sin}^{2} \frac{x}{\cos} ^ 2 x$,

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

$= {\cos}^{2} x$, as desired!