ɑ and ẞ are two positive acute angles and cos 2ɑ= (3cos2β-1)/(3-cos2β) Which of the following relation is correct?

Explanation:

Another Explanation (5): সমাধান: আমাদের দেওয়া আছে, \( \cos 2\alpha = \frac{3\cos 2\beta - 1}{3 - \cos 2\beta} \) আমরা জানি, \( \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \) এই সূত্র ব্যবহার করে পাই, \( \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{3\left(\frac{1 - \tan^2 \beta}{1 + \tan^2 \beta}\right) - 1}{3 - \left(\frac{1 - \tan^2 \beta}{1 + \tan^2 \beta}\right)} \) এখন, লব ও হরকে \( (1 + \tan^2 \beta) \) দিয়ে গুণ করে পাই, \( \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{3(1 - \tan^2 \beta) - (1 + \tan^2 \beta)}{3(1 + \tan^2 \beta) - (1 - \tan^2 \beta)} \) \( \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{3 - 3\tan^2 \beta - 1 - \tan^2 \beta}{3 + 3\tan^2 \beta - 1 + \tan^2 \beta} \) \( \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{2 - 4\tan^2 \beta}{2 + 4\tan^2 \beta} \) \( \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{1 - 2\tan^2 \beta}{1 + 2\tan^2 \beta} \) যোগ বিয়োগ প্রক্রিয়া করে পাই, \( \frac{(1 - \tan^2 \alpha) + (1 + \tan^2 \alpha)}{(1 - \tan^2 \alpha) - (1 + \tan^2 \alpha)} = \frac{(1 - 2\tan^2 \beta) + (1 + 2\tan^2 \beta)}{(1 - 2\tan^2 \beta) - (1 + 2\tan^2 \beta)} \) \( \frac{2}{-2\tan^2 \alpha} = \frac{2}{-4\tan^2 \beta} \) \( \frac{1}{\tan^2 \alpha} = \frac{1}{2\tan^2 \beta} \) \( \tan^2 \alpha = 2\tan^2 \beta \) অতএব, \( \tan \alpha = \sqrt{2} \tan \beta \) 🥳