Question

Prove that: $\frac{\sec \theta -1}{\sec \theta +1}={\tan }^2\frac{\theta }{2}$

Answer 1

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

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

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

We know that $\cos \theta =1-2{\sin }^2\frac{\theta }{2}$ and $\cos \theta =2{\cos }^2\frac{\theta }{2}-1$

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

$={\tan }^2\frac{\theta }{2}$