Question

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

Answer 1

LHS

$\frac{{\cos }^2\theta +{\tan }^2\theta -1}{{\sin }^2\theta }=\frac{{\cos }^2\theta }{{\sin }^2\theta }+\frac{{\tan }^2\theta }{{\sin }^2\theta }-\frac{1}{{\sin }^2\theta }={\cot }^2\theta +{\sec }^2\theta -{\csc }^2\theta ={\sec }^2\theta -1[\because {\csc }^2\theta -{\cot }^2\theta =1]={\tan }^2\theta [\because {\sec }^2\theta -{\tan }^2\theta =1]=RHS$

Hence, proved.