Prove that $ \frac{1 + tan^2 A}{1 + cot^2 A} = sec^2A -1 $

Prove that $ \frac{1 + tan^2 A}{1 + cot^2 A} = sec^2A -1 $

Answer

$$ LHS = \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \frac{\sin^2 A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^2A}} $$

$$ = \frac{\frac{\cos^2 A  +  \sin^2 A}{\cos^2A}} {\frac{\sin^2A + \cos^2A}{\sin^2A}} $$

$$ =  \frac{\frac{1}{\cos^2 A}} {\frac{1}{\sin2A}} = \frac{\sin^2A}{\cos^2A} = {\frac{1 - \cos^2A}{\cos^2A}} $$

$$ \frac{1}{\cos^2A} - 1 = \sec^2A -1 = \text{RHS} $$