Question

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

Answer 1

L.H.S. $={\cot }^2.\frac{\sec \theta -1}{1+\sin \theta }.\frac{\sec \theta +1}{\sec \theta +1}$
$=\frac{{\cot }^2\theta ({\sec }^2\theta -1)}{(\sec \theta +1)(1+\sin \theta )}$
$=\frac{{\cot }^2\theta {\tan }^2\theta }{(\sec \theta +1)}.\frac{1-\sin \theta }{1-{\sin }^2\theta }$
$=\frac{1}{1+\sec \theta }.\frac{1-\sin \theta }{{\cos }^2\theta }={\sec }^2\theta .\frac{1-\sin \theta }{1+sec\theta }$.