sinθ=1/√5 হলে, sin^-1(1/sqrt5)+tan^-1(1/3)=? 

দেওয়া আছে, \( \sin\theta = \frac{1}{\sqrt{5}} \)

সুতরাং, \( \theta = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \)

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

সুতরাং, \( \cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{1}{\sqrt{5}}\right)^2 = 1 - \frac{1}{5} = \frac{4}{5} \)

অতএব, \( \cos\theta = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \) (যেহেতু \( \theta \) প্রথম চতুর্ভাগে অবস্থিত)

এখন, \( \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \)

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

অতএব, \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = \tan^{-1}\left(\frac{1}{2}\right) \)

সুতরাং, \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \)

আমরা জানি, \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \)

অতএব, \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \)

আমরা জানি, \( \tan^{-1}(1) = \frac{\pi}{4} \)

সুতরাং, \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \) 🥳