\((\frac{1 + tan^2A}{1 + cot^2A}) = (\frac{1 - tanA}{1 - cotA})^2\) = tan^2A

Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

\((\frac{1 + tan^2A}{1 + cot^2A}) = (\frac{1 - tanA}{1 - cotA})^2\) = tan²A

Answer 1

LHS = \((\frac{1 + tan^2A}{1 + cot^2A}) = (\frac{1 - tanA}{1 - cotA})^2\)