cos θ=4/5 then,  (1-tan^2 θ)/(1+tan^2 θ)=? 

Explanation:

Another Explanation (5): ```html সমাধান: দেওয়া আছে, \( \cos \theta = \frac{4}{5} \). আমরা জানি, \( \sin^2 \theta + \cos^2 \theta = 1 \). সুতরাং, \( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \). তাহলে, \( \sin \theta = \pm \frac{3}{5} \). এখন, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\pm \frac{3}{5}}{\frac{4}{5}} = \pm \frac{3}{4} \). আমাদের নির্ণয় করতে হবে: \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \). \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - \left(\pm \frac{3}{4}\right)^2}{1 + \left(\pm \frac{3}{4}\right)^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{16 - 9}{16}}{\frac{16 + 9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{16} \times \frac{16}{25} = \frac{7}{25} \). অতএব, \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{7}{25} \). 🎉 ```