cosθ = 4/5 হলে, (1-tan^2θ)/(1+tan^2θ) = ? 

Explanation:

Another Explanation (5): দেওয়া আছে, \( cos\theta = \frac{4}{5} \) আমরা জানি, \( sin^2\theta + cos^2\theta = 1 \) তাহলে, \( sin^2\theta = 1 - cos^2\theta \) \( sin^2\theta = 1 - (\frac{4}{5})^2 \) \( sin^2\theta = 1 - \frac{16}{25} \) \( sin^2\theta = \frac{25-16}{25} \) \( sin^2\theta = \frac{9}{25} \) \( sin\theta = \pm\frac{3}{5} \) এখন, \( tan\theta = \frac{sin\theta}{cos\theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \) (আমরা ধনাত্মক মানটি নিলাম)। তাহলে, \( tan^2\theta = (\frac{3}{4})^2 = \frac{9}{16} \) এখন, \( \frac{1-tan^2\theta}{1+tan^2\theta} = \frac{1-\frac{9}{16}}{1+\frac{9}{16}} \) \( = \frac{\frac{16-9}{16}}{\frac{16+9}{16}} \) \( = \frac{\frac{7}{16}}{\frac{25}{16}} \) \( = \frac{7}{16} \times \frac{16}{25} \) \( = \frac{7}{25} \) অতএব, \( \frac{1-tan^2\theta}{1+tan^2\theta} = \frac{7}{25} \) 🥳