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,
\[
\boxed{\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{73}{161}\right)}
\]
