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

Jun 7, 2016

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

#### Explanation:

We use the Pythagorean identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\implies {\cos}^{2} \theta = 1 - {\sin}^{2} \theta$
$\implies \cos \theta = \sqrt{1 - {\sin}^{2} \theta}$
$\implies {\sec}^{2} \theta = \frac{1}{{\cos}^{2} \theta} = \frac{1}{1 - {\sin}^{2} \theta}$

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