Question

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

Answer 1

We need to prove $\frac{1+\sec \theta }{\sec \theta }=\frac{{\sin }^2\theta }{1-\cos \theta }$

Taking R. H. S. $=\frac{{\sin }^2\theta }{1-\cos \theta }$

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

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

$=1+\cos \theta$

Now taking L.H.S. $=\frac{1+\sec \theta }{\sec \theta }$

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

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

$=1+\cos \theta$

L.H.S. $=$ R.H.S.

Hence, it proved.