যদি tanθ + secθ = x হয় তবে sinθ =?

দেওয়া আছে: \( \tan\theta + \sec\theta = x \)

আমরা জানি, \( \sec^2\theta - \tan^2\theta = 1 \)

সুতরাং, \( (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1 \)

যেহেতু \( \sec\theta + \tan\theta = x \), তাই \( x(\sec\theta - \tan\theta) = 1 \)

অতএব, \( \sec\theta - \tan\theta = \frac{1}{x} \)

এখন, \( \sec\theta + \tan\theta = x \) এবং \( \sec\theta - \tan\theta = \frac{1}{x} \) যোগ করে পাই,

\( 2\sec\theta = x + \frac{1}{x} = \frac{x^2 + 1}{x} \)

সুতরাং, \( \sec\theta = \frac{x^2 + 1}{2x} \)

আবার, \( \sec\theta + \tan\theta = x \) থেকে \( \sec\theta - \tan\theta = \frac{1}{x} \) বিয়োগ করে পাই,

\( 2\tan\theta = x - \frac{1}{x} = \frac{x^2 - 1}{x} \)

সুতরাং, \( \tan\theta = \frac{x^2 - 1}{2x} \)

আমরা জানি, \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) এবং \( \sec\theta = \frac{1}{\cos\theta} \)

সুতরাং, \( \sin\theta = \tan\theta \cdot \cos\theta = \frac{\tan\theta}{\sec\theta} \)

\( \sin\theta = \frac{\frac{x^2 - 1}{2x}}{\frac{x^2 + 1}{2x}} = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} \)

অতএব, \( \sin\theta = \frac{x^2 - 1}{x^2 + 1} \) 🎉