tan^-1(1/7)+tan^-1(1/7)+tan^-1(1/7)= ? 

Another Explanation (5): প্রশ্নঃ \(\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{7}\right) = ?\) সমাধানঃ ধরা যাক, \(A = \tan^{-1}\left(\frac{1}{7}\right)\) তাহলে, \[ \text{প্রথম দুটি যোগফলঃ} \quad A + A = 2A \] এবং, \[ \text{তৃতীয় যোগফলঃ} \quad 2A + A = 3A \] অর্থাৎ, মূল প্রশ্নটি হলো: \[ 3A = 3 \times \tan^{-1}\left(\frac{1}{7}\right) \] তাই, \[ \text{আমরা চাই } \tan(3A) \] তাহলে, \[ \tan(3A) = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} \] এখানে, \(\tan A = \frac{1}{7}\) সুতরাং, \[ \tan(3A) = \frac{3 \times \frac{1}{7} - \left(\frac{1}{7}\right)^3}{1 - 3 \times \left(\frac{1}{7}\right)^2} \] গণনা করি, উপরের অংশ: \[ 3 \times \frac{1}{7} = \frac{3}{7} \] \[ \left(\frac{1}{7}\right)^3 = \frac{1}{343} \] তাহলে, \[ \text{উপরে} = \frac{3}{7} - \frac{1}{343} \] কমন ডেনোমিনেটর 343: \[ \frac{3}{7} = \frac{3 \times 49}{7 \times 49} = \frac{147}{343} \] অতএব, \[ \text{উপরে} = \frac{147}{343} - \frac{1}{343} = \frac{146}{343} \] নিচে: \[ 1 - 3 \times \left(\frac{1}{7}\right)^2 = 1 - 3 \times \frac{1}{49} = 1 - \frac{3}{49} \] \[ = \frac{49}{49} - \frac{3}{49} = \frac{46}{49} \] তাহলে, \[ \tan(3A) = \frac{\frac{146}{343}}{\frac{46}{49}} = \frac{146}{343} \times \frac{49}{46} \] সরলীকরণ: \[ = \frac{146 \times 49}{343 \times 46} \] নোট করি, \[ 146 = 2 \times 73 \] \[ 46 = 2 \times 23 \] \[ 343 = 7 \times 7 \times 7 \] অতএব, \[ \tan(3A) = \frac{2 \times 73 \times 49}{7 \times 7 \times 7 \times 2 \times 23} \] সংক্ষিপ্ত করে, \[ = \frac{73 \times 49}{7^3 \times 23} \] আরো সরলীকরণ: \[ 49 = 7 \times 7 \] অতএব, \[ \tan(3A) = \frac{73 \times 7 \times 7}{7^3 \times 23} = \frac{73 \times 7 \times 7}{7 \times 7 \times 7 \times 23} \] একটি 7 ও এক 7 কেটে গেলে, \[ = \frac{73}{7 \times 23} = \frac{73}{161} \] অতএব, \[ 3A = \tan^{-1}\left(\frac{73}{161}\right) \] সুতরাং, \[ \boxed{\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{73}{161}\right)} \]