# What is tan^4theta/(1-sin^2theta in terms of non-exponential trigonometric functions?

Dec 17, 2015

$\frac{2}{\cos 2 x - 1}$

#### Explanation:

$f \left(x\right) = {\tan}^{4} \frac{x}{1 - {\sin}^{2} x} = \left({\sin}^{4} \frac{x}{\cos} ^ 4 x\right) \left(\frac{1}{\cos} ^ 2 x\right)$ =
$= \left({\sin}^{4} \frac{x}{\cos} ^ 6 x\right)$
Develop ${\sin}^{4} x$ and ${\cos}^{6} x$

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

${\cos}^{6} x = \left({\cos}^{2} x\right) \left({\cos}^{2} x\right) \left({\cos}^{2} x\right) =$
$= \left(\frac{1}{8}\right) \left(\cos 2 x - 1\right) \left(\cos 2 x - 1\right) \left(\cos 2 x - 1\right)$
Finally,
$f \left(x\right) = \frac{\left(\frac{1}{4}\right) {\left(\cos 2 x - 1\right)}^{2}}{\left(\frac{1}{8}\right) {\left(\cos 2 x - 1\right)}^{3}} =$
$= \frac{2}{\cos 2 x - 1} .$