Question

1+sin2a/1-sin2a=(1+tana/1-tana)^2​

Answer 1

`\Large \mathbb{SOLUTION:}`

`\begin{array}{l} (1 + \sin 2A)/(1 - \sin 2A) = ((1 + \tan A)^2)/((1 - \tan A)^2) \\ \\ (1 + 2\sin A\cos A)/(1 - 2\sin A\cos A) = ((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. \\ \\ ((1 + 2\sin A\cos A)/(\cos^2 A))/((1 - 2\sin A\cos A)/(\cos^2 A)) = ((1 + \tan A)^2)/((1 - \tan A)^2) \\ \\ ((1)/(\cos^2 A) + (2\sin A\cos A)/(\cos^2 A))/((1)/(\cos^2 A) - (2\sin A\cos A)/(\cos^2 A)) = ((1 + \tan A)^2)/((1 - \tan A)^2)\\ \\ \frac{\sec^2 A + 2\tan A} {\sec^2 A- 2\tan A} = ((1 + \tan A)^2)/((1 - \tan A)^2) \\ \\ \frac{1 + \tan^2 A + 2\tan A} {1 + \tan^2 A - 2\tan A} = ((1 + \tan A)^2)/((1 - \tan A)^2) \\ \\ \because \sec^2 A = 1 + \tan^2 A\ (\text{Pythagorean Identity}) \\ \\ \text{Rearranging, we get} \\ \\ \frac{\tan^2 A + 2\tan A + 1} {\tan^2 A - 2\tan A + 1} = ((1 + \tan A)^2)/((1 - \tan A)^2) \\ \\ ((1 + \tan A)^2)/((1 - \tan A)^2) = ((1 + \tan A)^2)/((1 - \tan A)^2)\\ \\ \text{LHS} = \text{RHS}_{\boxed{\:}}\end{array}`