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

Aug 2, 2016

${\tan}^{2} \frac{\theta}{1 - {\sin}^{4} \theta} - {\csc}^{2} \theta$

$= {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta \cdot \frac{1}{\left(1 - {\sin}^{2} \theta\right) \left(1 + {\sin}^{2} \theta\right)} - {\csc}^{2} \theta$

$= {\sin}^{2} \frac{\theta}{\cos} ^ 4 \theta \cdot \frac{1}{1 + {\sin}^{2} \theta} - {\csc}^{2} \theta$

$= \frac{8 {\sin}^{2} \theta}{2 {\cos}^{2} \theta} ^ 2 \cdot \frac{1}{2 + 2 {\sin}^{2} \theta} - {\csc}^{2} \theta$

$= \frac{4 \left(1 - \cos 2 \theta\right)}{1 + \cos 2 \theta} ^ 2 \cdot \frac{1}{2 + 1 - \cos 2 \theta} - {\csc}^{2} \theta$

$= \frac{8 \left(1 - \cos 2 \theta\right)}{2 + 4 \cos 2 \theta + 2 {\cos}^{2} 2 \theta} \cdot \frac{1}{3 - \cos 2 \theta} - \frac{2}{2 {\sin}^{2} \theta}$

$= \frac{8 \left(1 - \cos 2 \theta\right)}{3 + 4 \cos 2 \theta + \cos 4 \theta} \cdot \frac{1}{3 - \cos 2 \theta} - \frac{2}{1 - \cos 2 \theta}$

Aug 2, 2016

${\tan}^{2} \frac{\theta}{1 - {\sin}^{4} \theta} - {\csc}^{2} \theta$

$= {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta \cdot \frac{1}{\left(1 - {\sin}^{2} \theta\right) \left(1 + {\sin}^{2} \theta\right)} - {\csc}^{2} \theta$

$= {\sin}^{2} \frac{\theta}{\cos} ^ 4 \theta \cdot \frac{1}{1 + {\sin}^{2} \theta} - {\csc}^{2} \theta$

$= \frac{8 {\sin}^{2} \theta}{2 {\cos}^{2} \theta} ^ 2 \cdot \frac{1}{2 + 2 {\sin}^{2} \theta} - {\csc}^{2} \theta$

$= \frac{4 \left(1 - \cos 2 \theta\right)}{1 + \cos 2 \theta} ^ 2 \cdot \frac{1}{2 + 1 - \cos 2 \theta} - {\csc}^{2} \theta$

$= \frac{8 \left(1 - \cos 2 \theta\right)}{2 + 4 \cos 2 \theta + 2 {\cos}^{2} 2 \theta} \cdot \frac{1}{3 - \cos 2 \theta} - \frac{2}{2 {\sin}^{2} \theta}$

$= \frac{8 \left(1 - \cos 2 \theta\right)}{3 + 4 \cos 2 \theta + \cos 4 \theta} \cdot \frac{1}{3 - \cos 2 \theta} - \frac{2}{1 - \cos 2 \theta}$