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

দেওয়া আছে, \( \cos\theta = \frac{4}{5} \)।

আমরা জানি, \( \sin^2\theta + \cos^2\theta = 1 \)।

সুতরাং, \( \sin^2\theta = 1 - \cos^2\theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \)।

অতএব, \( \sin\theta = \pm\frac{3}{5} \)।

এখন, \( \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\pm\frac{3}{5}}{\frac{4}{5}} = \pm\frac{3}{4} \)।

তাহলে, \( \tan^2\theta = \left(\pm\frac{3}{4}\right)^2 = \frac{9}{16} \)।

এখন, \( \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{16-9}{16}}{\frac{16+9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{16} \times \frac{16}{25} = \frac{7}{25} \)। 🎉

সুতরাং, \( \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{7}{25} \)।🥳