f(x) = tan^-1x হলে  tan{f(1/2)-f(1/5)} = কত?

Another Explanation (5): প্রশ্ন: \(f(x) = \tan^{-1} x\), তাহলে \(\tan\left(f\left(\frac{1}{2}\right) - f\left(\frac{1}{5}\right)\right) = \ কত?\) সমাধান: প্রথমে, আমাদের জানা দরকার যে: \[ f(x) = \tan^{-1} x \] অর্থাৎ: \[ f\left(\frac{1}{2}\right) = \tan^{-1} \frac{1}{2} \quad \text{এবং} \quad f\left(\frac{1}{5}\right) = \tan^{-1} \frac{1}{5} \] আমরা জানি যে: \[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \] এখানে, \(A = f\left(\frac{1}{2}\right)\) এবং \(B = f\left(\frac{1}{5}\right)\), ফলে: \[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \] যেহেতু \(\tan A = \frac{1}{2}\) এবং \(\tan B = \frac{1}{5}\), তাহলে: \[ \tan\left(f\left(\frac{1}{2}\right) - f\left(\frac{1}{5}\right)\right) = \frac{\frac{1}{2} - \frac{1}{5}}{1 + \frac{1}{2} \times \frac{1}{5}} \] গণনা করি: \[ \text{উপরের অংশ:} \quad \frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \] নিচের অংশ: \[ 1 + \frac{1}{2} \times \frac{1}{5} = 1 + \frac{1}{10} = \frac{10}{10} + \frac{1}{10} = \frac{11}{10} \] অতএব: \[ \tan\left(f\left(\frac{1}{2}\right) - f\left(\frac{1}{5}\right)\right) = \frac{\frac{3}{10}}{\frac{11}{10}} = \frac{3}{10} \times \frac{10}{11} = \frac{3}{11} \] অতএব, উত্তর হলো: $3/11$