# Question #cc719

Jul 24, 2016

${\sin}^{2} 2 x \left({\cot}^{2} x - {\tan}^{2} x\right) = 4 \cos 2 x$
$= \sin 2 x \cdot \sin 2 x \left({\cot}^{2} x - {\tan}^{2} x\right)$

Using the double angle formula for sin and converting cot and tan to terms with sin and cos we now have

$= 4 {\sin}^{2} x {\cos}^{2} x \left({\cos}^{2} \frac{x}{\sin} ^ 2 x - {\sin}^{2} \frac{x}{\cos} ^ 2 x\right)$
$= 4 {\sin}^{2} x {\cos}^{2} x \left(\frac{{\cos}^{4} x - {\sin}^{4} x}{{\sin}^{2} x {\cos}^{2} x}\right)$
$= 4 {\sin}^{2} x {\cos}^{2} x \frac{\left({\cos}^{2} x + {\sin}^{2} x\right) \left({\cos}^{2} x - {\sin}^{2} x\right)}{\left({\sin}^{2} x {\cos}^{2} x\right)}$
$= 4 {\sin}^{2} x {\cos}^{2} x \left(\frac{{\cos}^{2} x - {\sin}^{2} x}{{\sin}^{2} x \cdot {\cos}^{2} x}\right)$ since ${\cos}^{2} x + {\sin}^{2} x = 1$
$= \frac{4 {\cos}^{4} {\sin}^{2} x - {\sin}^{4} x {\cos}^{2} x}{{\sin}^{2} x {\cos}^{2} x}$
$= \frac{4 \cancel{{\sin}^{2} x {\cos}^{2} x} \left({\cos}^{2} x - {\sin}^{2} x\right)}{\cancel{{\sin}^{2} x {\cos}^{2} x}}$
$= 4 \left({\cos}^{2} x - {\sin}^{2} x\right)$
$= 4 \cos 2 x$ since ${\cos}^{2} x - {\sin}^{2} x = \cos 2 x$