\( \cos \theta = \frac{2x}{x^2+1} \), \( x>1 \) হলে \( \tan \theta + \sec \theta\) =?

দেওয়া আছে, \( \cos \theta = \frac{2x}{x^2+1} \) এবং \( x > 1 \)। আমাদের \( \tan \theta + \sec \theta \) এর মান নির্ণয় করতে হবে।

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

সুতরাং, \( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{2x}{x^2+1}\right)^2 \)

\( \sin^2 \theta = 1 - \frac{4x^2}{(x^2+1)^2} = \frac{(x^2+1)^2 - 4x^2}{(x^2+1)^2} \)

\( \sin^2 \theta = \frac{x^4 + 2x^2 + 1 - 4x^2}{(x^2+1)^2} = \frac{x^4 - 2x^2 + 1}{(x^2+1)^2} \)

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

অতএব, \( \sin \theta = \pm \frac{x^2-1}{x^2+1} \)। যেহেতু \( x > 1 \), তাই \( x^2 > 1 \) এবং \( x^2 - 1 > 0 \)। \( \theta \) প্রথম চতুর্ভাগে অবস্থিত। সুতরাং, \( \sin \theta \) ধনাত্মক হবে।

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

এখন, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{x^2-1}{x^2+1}}{\frac{2x}{x^2+1}} = \frac{x^2-1}{2x} \)

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

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

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

সুতরাং, \( \tan \theta + \sec \theta = x \) 🥳