How do you simplify (1 + sec theta)/sec theta = sin^2 theta/(1 - cos theta)?

Nov 27, 2015

Explanation:

Remember: The trigonometric identities '

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

$\frac{1}{\cos} \theta = \sec \theta$ $\Leftrightarrow$ $\frac{1}{\sec} \theta = \cos \theta$

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

$\implies$ If we start from the right hand side, multiply by the conjugate of the denominator

(1+sectheta)/sectheta = [(sin^2theta)/(1-costheta)]color(red)([(1+costheta)/(1+costheta)]

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

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

(1+sectheta)/sectheta = [cancel(sin^2theta) (1+costheta)]/(cancel(sin^2theta)

$\frac{1 + \sec \theta}{\sec} \theta = 1 + \cos \theta$

$\frac{1 + \sec \theta}{\sec} \theta = 1 + \frac{1}{\sec \theta}$ common denominator

$\frac{1 + \sec \theta}{\sec} \theta = 1 \cdot \frac{\textcolor{red}{\sec} \theta}{\textcolor{red}{\sec}} \theta + \frac{1}{\sec \theta}$

$\frac{1 + \sec \theta}{\sec} \theta = \frac{1 + \sec \theta}{\sec} \theta$

Done!