Question

$\frac{\cos 2A}{1-\sin 2A}=\frac{1+\tan A}{1-\tan A}$

Answer 1

Solving LHS, We get

$\frac{cos2A}{1-sin2A}$

$⟹\frac{\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}}{1-\frac{2tanA}{1+ta{n}^{2}A}}$

$⟹\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A-2tanA}$

$⟹\frac{(1-tanA)(1+tanA)}{(1-tanA{)}^2}$

$⟹\frac{(1+tanA)}{(1-tanA)}$ = RHS

Hence proved.