যদি cosA=4/5 হয়, তবে  (1+tan^2A)/(1-tan^2A) এর মান-

Explanation:

Another Explanation (5): ```html যদি \( \cos A = \frac{4}{5} \) হয়, তবে \( \frac{1 + \tan^2 A}{1 - \tan^2 A} \) এর মান নির্ণয় করতে হবে। 🤔 আমরা জানি, \( \sin^2 A + \cos^2 A = 1 \) সুতরাং, \( \sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} \) অতএব, \( \sin A = \sqrt{\frac{9}{25}} = \frac{3}{5} \) (যেহেতু \( A \) একটি কোণ, তাই \( \sin A \) ধনাত্মক হবে)। এখন, \( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \) 🤓 তাহলে, \( \tan^2 A = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \) সুতরাং, \( \frac{1 + \tan^2 A}{1 - \tan^2 A} = \frac{1 + \frac{9}{16}}{1 - \frac{9}{16}} = \frac{\frac{16 + 9}{16}}{\frac{16 - 9}{16}} = \frac{\frac{25}{16}}{\frac{7}{16}} = \frac{25}{16} \times \frac{16}{7} = \frac{25}{7} \) 🎉 অতএব, \( \frac{1 + \tan^2 A}{1 - \tan^2 A} = \frac{25}{7} \) 😎 ```