# How do you simplify sin^2theta-cos^2theta+tan^2theta to non-exponential trigonometric functions?

Jul 8, 2016

The given expression

$= {\sin}^{2} \theta - {\cos}^{2} \theta + {\tan}^{2} \theta$

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

Now recalling the identities

$\left({\cos}^{2} \theta - {\sin}^{2} \theta\right) = \cos 2 \theta$

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

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

The given expression becomes

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

$= \frac{- \cos 2 \theta - {\cos}^{2} 2 \theta + 1 - \cos 2 \theta}{1 + \cos 2 \theta}$

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

$= \frac{\frac{1}{2} - 2 \cos 2 \theta - \frac{1}{2} \cos 4 \theta}{1 + \cos 2 \theta}$

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