tan^-1 1/2 +tan^-1 1/3 কত?

Another Explanation (5): প্রশ্ন: \(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}\) কত? উত্তর: \(\frac{\pi}{4}\) সমাধান: আমরা জানি, \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right), \quad \text{যদি } 1 - ab \neq 0 \] এখানে, \(a = \frac{1}{2}\), \(b = \frac{1}{3}\) তাহলে, \[ \begin{aligned} \theta &= \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} \\ &= \tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} \right) \\ &= \tan^{-1} \left( \frac{\frac{3}{6} + \frac{2}{6}}{1 - \frac{1}{6}} \right) \\ &= \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) \\ &= \tan^{-1}(1) \\ &= \frac{\pi}{4} \end{aligned} \] অতএব, \[ \boxed{\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \frac{\pi}{4}} \]