(1-tan^2(45°-A))/(1+tan^2(45°-A))=? 

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \( \tan(45^\circ - A) = \frac{\tan 45^\circ - \tan A}{1 + \tan 45^\circ \tan A} = \frac{1 - \tan A}{1 + \tan A} \) 🤓 তাহলে, \( \frac{1 - \tan^2(45^\circ - A)}{1 + \tan^2(45^\circ - A)} = \frac{1 - (\frac{1 - \tan A}{1 + \tan A})^2}{1 + (\frac{1 - \tan A}{1 + \tan A})^2} \) 🤔 \( = \frac{(1 + \tan A)^2 - (1 - \tan A)^2}{(1 + \tan A)^2 + (1 - \tan A)^2} \) 🤯 \( = \frac{(1 + 2\tan A + \tan^2 A) - (1 - 2\tan A + \tan^2 A)}{(1 + 2\tan A + \tan^2 A) + (1 - 2\tan A + \tan^2 A)} \) 🤩 \( = \frac{1 + 2\tan A + \tan^2 A - 1 + 2\tan A - \tan^2 A}{1 + 2\tan A + \tan^2 A + 1 - 2\tan A + \tan^2 A} \) 🥳 \( = \frac{4\tan A}{2 + 2\tan^2 A} \) 😲 \( = \frac{2\tan A}{1 + \tan^2 A} \) 😎 আমরা জানি, \( \sin 2A = \frac{2\tan A}{1 + \tan^2 A} \) 🥰 সুতরাং, \( \frac{1 - \tan^2(45^\circ - A)}{1 + \tan^2(45^\circ - A)} = \sin 2A \) 🎉