Question

prove that: ${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta {\sin }^2\theta$

Answer 1

$LHS={\tan }^2\theta -{\sin }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta ={\sin }^2\theta ({\sec }^2\theta -1)={\sin }^2\theta {\tan }^2\theta =RHS$