2tan^-1 (1/5) - tan^-1 (1/3) =tan^-1x হলেx এর মান কত ?

Explanation:

Another Explanation (5): bài toán: \(2\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}x\) হলে \(x\) এর মান নির্ণয়। সমাধান: আমরা জানি, \(2\tan^{-1}a = \tan^{-1}\left(\frac{2a}{1-a^2}\right)\) অতএব, \(2\tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2}\right)\) \(= \tan^{-1}\left(\frac{\frac{2}{5}}{1 - \frac{1}{25}}\right)\) \(= \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)\) \(= \tan^{-1}\left(\frac{2}{5} \cdot \frac{25}{24}\right)\) \(= \tan^{-1}\left(\frac{5}{12}\right)\) এখন, \(2\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{5}{12}\right) - \tan^{-1}\left(\frac{1}{3}\right)\) আমরা জানি, \(\tan^{-1}a - \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)\) সুতরাং, \(\tan^{-1}\left(\frac{5}{12}\right) - \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{5}{12} - \frac{1}{3}}{1 + \frac{5}{12} \cdot \frac{1}{3}}\right)\) \(= \tan^{-1}\left(\frac{\frac{5-4}{12}}{1 + \frac{5}{36}}\right)\) \(= \tan^{-1}\left(\frac{\frac{1}{12}}{\frac{36+5}{36}}\right)\) \(= \tan^{-1}\left(\frac{\frac{1}{12}}{\frac{41}{36}}\right)\) \(= \tan^{-1}\left(\frac{1}{12} \cdot \frac{36}{41}\right)\) \(= \tan^{-1}\left(\frac{3}{41}\right)\) অতএব, \(\tan^{-1}x = \tan^{-1}\left(\frac{3}{41}\right)\) সুতরাং, \(x = \frac{3}{41}\) 🎉🎉🎉