Question

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

Answer 1

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

$=1+\cos \theta$

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

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

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

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