tan^-1(3/5)+tan^-1(3/4)=?

Another Explanation (5): প্রশ্ন: \(\tan^{-1}\left(\frac{3}{5}\right) + \tan^{-1}\left(\frac{3}{4}\right) = ?\) উত্তর: \(\tan^{-1}\left(\frac{27}{11}\right)\) সমাধান: আমরা জানি, \[ \text{যদি } \alpha = \tan^{-1} a \text{ এবং } \beta = \tan^{-1} b, \text{ তবে } \alpha + \beta = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \] যখন \(1 - ab \neq 0\). এখানে, \[ a = \frac{3}{5}, \quad b = \frac{3}{4} \] তাহলে, \[ \alpha + \beta = \tan^{-1} \left( \frac{\frac{3}{5} + \frac{3}{4}}{1 - \left(\frac{3}{5} \times \frac{3}{4}\right)} \right) \] প্রথমে, উপরের অংশ: \[ \frac{3}{5} + \frac{3}{4} = \frac{(3 \times 4) + (3 \times 5)}{5 \times 4} = \frac{12 + 15}{20} = \frac{27}{20} \] নিচের অংশ: \[ 1 - \left(\frac{3}{5} \times \frac{3}{4}\right) = 1 - \left(\frac{9}{20}\right) = \frac{20}{20} - \frac{9}{20} = \frac{11}{20} \] অতএব, \[ \alpha + \beta = \tan^{-1} \left( \frac{\frac{27}{20}}{\frac{11}{20}} \right) = \tan^{-1} \left( \frac{27}{20} \times \frac{20}{11} \right) = \tan^{-1} \left( \frac{27}{11} \right) \] অতএব, \[ \boxed{\tan^{-1} \left( \frac{27}{11} \right)} \]