Question

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

Answer 1

First, we consider $\frac{1+\sec \theta }{\sec \theta }$

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

$=1+\cos \theta$

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

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

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