Question

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

Answer 1

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

Question can be rewritten as : $\frac{1-\cos \theta }{1+\cos \theta }$

Multiply and divide by $(1+\cos \theta )$

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

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

$={\left(\frac{\sin \theta }{1+\cos \theta }\right)}^2$

$=RHS$

Hence proved