# Show that  sin^2x tan^2x = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) ) ?

Oct 21, 2017

We will utilise the following identities:

$\cos 2 A \equiv 2 {\cos}^{2} A - 1 \implies {\cos}^{2} A = \frac{1}{2} \left(1 + \cos 2 A\right)$
${\sin}^{2} A + {\cos}^{2} A \equiv 1 \setminus \setminus \setminus \implies {\sin}^{2} A = \frac{1}{2} \left(1 - \cos 2 A\right)$

Then:

${\sin}^{2} x {\tan}^{2} x \equiv {\sin}^{2} x {\sin}^{2} \frac{x}{\cos} ^ 2 x$

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

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

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

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

$\text{ } = \frac{1 - 2 \cos 2 x + \frac{1}{2} \left(1 + \cos 4 x\right)}{2 \left(1 + \cos 2 x\right)}$

$\text{ } = \frac{\frac{1}{2} \left(2 - 4 \cos 2 x + 1 + \cos 4 x\right)}{2 \left(1 + \cos 2 x\right)}$
$\text{ } = \frac{3 - 4 \cos 2 x + \cos 4 x}{4 \left(1 + \cos 2 x\right)} \setminus \setminus \setminus$ QED