tanθ + secθ=x হলে, cosecθ- এর মান কত? 

দেওয়া আছে, \( \tan \theta + \sec \theta = x \) 🤔

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

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

বা, \( x(\sec \theta - \tan \theta) = 1 \)

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

এখন, \( \tan \theta + \sec \theta = x \) 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, \( \tan \theta + \sec \theta = x \) 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, \( \sin \theta = \frac{\tan \theta}{\sec \theta} \)

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

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