\( \tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{3}\right) \) এর মান কো???টি?

Another Explanation (5): প্রশ্ন: \( \tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{3}\right) \) এর মান কোনটি? উত্তর: \( \frac{\pi}{4} \) সমাধান: আমরা জানি, \[ \arctan a + \arctan b = \arctan \left( \frac{a + b}{1 - ab} \right) \] যদি \( 1 - ab \neq 0 \)। এখানে, \( a = \frac{1}{2} \), \( b = \frac{1}{3} \)। তাহলে, \[ \arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{1}{3}\right) = \arctan \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} \right) \] গণনা করি, \[ \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{3}{6} + \frac{2}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \] অতএব, \[ \arctan \left( 1 \right) = \frac{\pi}{4} \] অতএব, \[ \boxed{ \arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{1}{3}\right) = \frac{\pi}{4} } \]