Question

Prove that: $\frac{{\text{cosec}}^2\theta -{\cos }^2\theta }{{\cot }^2\theta }={\sec }^2\theta -{\sin }^2\theta$

Answer 1

$LHS=\frac{{\text{cosec}}^2\theta -{\cos }^2\theta }{{\cot }^2\theta }$

$=\frac{{\text{cosec}}^2\theta }{{\cot }^2\theta }-\frac{{\cos }^2\theta }{{\cot }^2\theta }$

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

$={\text{sec}}^2\theta -{\sin }^2\theta$

$=RHS$

Hence Proved