[tex] \Large \mathbb{SOLUTION:} [/tex]
[tex] \begin{array}{l} \dfrac{1 + \sin 2A}{1 - \sin 2A} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} \\ \\ \dfrac{1 + 2\sin A\cos A}{1 - 2\sin A\cos A} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} \\ \\ \because \sin 2A = 2\sin A\cos A\ (\text{Double Angle Identity}) \\ \\ \text{Divide both numerator and denominator of} \\ \text{LHS by }\cos^2 A. \\ \\ \dfrac{\frac{1 + 2\sin A\cos A}{\cos^2 A}}{\frac{1 - 2\sin A\cos A}{\cos^2 A}} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} \\ \\ \dfrac{\frac{1}{\cos^2 A} + \frac{2\sin A\cos A}{\cos^2 A}}{\frac{1}{\cos^2 A} - \frac{2\sin A\cos A}{\cos^2 A}} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2}\\ \\ \dfrac{\sec^2 A + 2\tan A} {\sec^2 A- 2\tan A} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} \\ \\ \dfrac{1 + \tan^2 A + 2\tan A} {1 + \tan^2 A - 2\tan A} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} \\ \\ \because \sec^2 A = 1 + \tan^2 A\ (\text{Pythagorean Identity}) \\ \\ \text{Rearranging, we get} \\ \\ \dfrac{\tan^2 A + 2\tan A + 1} {\tan^2 A - 2\tan A + 1} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} \\ \\ \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2} = \dfrac{(1 + \tan A)^2}{(1 - \tan A)^2}\\ \\ \text{LHS} = \text{RHS}_{\boxed{\:}}\end{array} [/tex]
