Question

Solve it:-
show that $\frac{1+\sec \theta }{\sec \theta }=\frac{{\sin }^2\theta }{1-\cos \theta }$

Answer 1

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

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

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

$=1+\cos \theta$

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

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

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

Hence proved.