Question

Prove that $(\tan \alpha {\sec }^2\alpha +\cot \alpha co{\sec }^2\alpha )-(\tan \alpha +\cot \alpha )=2co\sec 2\alpha [{\sec }^2\alpha +co{\sec }^2\alpha -3]$

Answer 1

LHS $=(\frac{\sin \alpha }{{\cos }^3\alpha }+\frac{\cos \alpha }{{\sin }^3\alpha })-\frac{2}{\sin 2\alpha }=\frac{1-2{\sin }^2\alpha {\cos }^2\alpha }{{\sin }^2\alpha {\cos }^2\alpha }.\frac{2}{2\sin \alpha \cos \alpha }-\frac{2}{\sin 2\alpha }=2co\sec 2\alpha [\frac{{\sin }^2\alpha +{\cos }^2\alpha }{{\sin }^2\alpha {\cos }^2\alpha }-3]$

$=2cosec2\alpha [{\sec }^2\alpha +cose{c}^2\alpha -3]$=Rhs

LHS=RHS

Hence proof