# How do you express sin theta - cot^2 theta + tan^2 theta  in terms of cos theta ?

Jan 31, 2016

here is how,

#### Explanation:

$\sin \theta - {\cot}^{2} \theta + {\tan}^{2} \theta$

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

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

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

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

=(cancel(cos^2theta)-cos^4theta+1-cos^4thetacancel(-cos^2theta))/(cos^2thetasqrt(1-cos^2theta)

$= \frac{1 - 2 {\cos}^{4} \theta}{\sqrt{{\cos}^{4} \theta - {\cos}^{6} \theta}}$