tanθ+secθ=x হলে, sinθ এর মান কত?

Explanation:

Another Explanation (5): দেওয়া আছে, \( \tan \theta + \sec \theta = x \) আমরা জানি, \( \sec^2 \theta - \tan^2 \theta = 1 \) \(\implies (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \) \(\implies x (\sec \theta - \tan \theta) = 1 \) \(\implies \sec \theta - \tan \theta = \frac{1}{x} \) এখন, \( \tan \theta + \sec \theta = x \) এবং \( \sec \theta - \tan \theta = \frac{1}{x} \) সমীকরণ দুটি যোগ করে পাই, \( 2 \sec \theta = x + \frac{1}{x} \) \(\implies 2 \sec \theta = \frac{x^2 + 1}{x} \) \(\implies \sec \theta = \frac{x^2 + 1}{2x} \) \(\implies \cos \theta = \frac{2x}{x^2 + 1} \) আবার, \( \tan \theta + \sec \theta = x \) থেকে \( \sec \theta - \tan \theta = \frac{1}{x} \) বিয়োগ করে পাই, \( 2 \tan \theta = x - \frac{1}{x} \) \(\implies 2 \tan \theta = \frac{x^2 - 1}{x} \) \(\implies \tan \theta = \frac{x^2 - 1}{2x} \) আমরা জানি, \( \sin \theta = \tan \theta \cdot \cos \theta \) \(\implies \sin \theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} \) \(\implies \sin \theta = \frac{x^2 - 1}{x^2 + 1} \) অতএব, \( \sin \theta = \frac{x^2 - 1}{x^2 + 1} \) ✅