\( \tan^{-1} \left( \frac{1}{7} \right) + \tan^{-1} \left( \frac{1}{8} \right) + \tan^{-1} \left( \frac{1}{18} \right) = ? \)

প্রথমে দেওয়া সমীকরণটি হলো:

\[ \tan^{-1} \left( \frac{1}{7} \right) + \tan^{-1} \left( \frac{1}{8} \right) + \tan^{-1} \left( \frac{1}{18} \right) \]

আমরা জানি,

\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - a b} \right), \quad \text{যখন} \quad 1 - a b \neq 0. \]

প্রথম ধাপে, প্রথম দুই টার্ন্টের যোগফল নির্ণয় করি:

\[ A = \tan^{-1} \left( \frac{1}{7} \right), \quad B = \tan^{-1} \left( \frac{1}{8} \right) \]

তাহলে,

\[ A + B = \tan^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) \]

সুতরাং, numerator:

\[ \frac{1}{7} + \frac{1}{8} = \frac{8}{56} + \frac{7}{56} = \frac{15}{56} \]

এবং denominator:

\[ 1 - \frac{1}{7} \times \frac{1}{8} = 1 - \frac{1}{56} = \frac{56}{56} - \frac{1}{56} = \frac{55}{56} \]

অতএব,

\[ A + B = \tan^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan^{-1} \left( \frac{15}{56} \times \frac{56}{55} \right) = \tan^{-1} \left( \frac{15}{55} \right) = \tan^{-1} \left( \frac{3}{11} \right) \]

এখন, শেষ টার্ন্ট যোগ করি:

\[ C = \tan^{-1} \left( \frac{1}{18} \right) \]

সুতরাং, সমাধান হবে:

\[ (A + B) + C = \tan^{-1} \left( \frac{3}{11} \right) + \tan^{-1} \left( \frac{1}{18} \right) \]

আবার, যোগফল নির্ণয় করি:

\[ D = \tan^{-1} \left( \frac{3}{11} \right) + \tan^{-1} \left( \frac{1}{18} \right) = \tan^{-1} \left( \frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \times \frac{1}{18}} \right) \]

\[ \frac{3}{11} + \frac{1}{18} = \frac{3 \times 18}{11 \times 18} + \frac{1 \times 11}{18 \times 11} = \frac{54}{198} + \frac{11}{198} = \frac{65}{198} \]

\[ 1 - \frac{3}{11} \times \frac{1}{18} = 1 - \frac{3}{198} = 1 - \frac{1}{66} = \frac{66}{66} - \frac{1}{66} = \frac{65}{66} \]

\[ D = \tan^{-1} \left( \frac{\frac{65}{198}}{\frac{65}{66}} \right) = \tan^{-1} \left( \frac{65}{198} \times \frac{66}{65} \right) = \tan^{-1} \left( \frac{66}{198} \right) = \tan^{-1} \left( \frac{1}{3} \right) \] এখানে, \(\tan^{-1} \left( \frac{1}{3} \right)\) এর সাথে সম্পর্কিত, আমরা জানি: \[ \cot^{-1} 3 = \tan^{-1} \left( \frac{1}{3} \right) \] অর্থাৎ, \[ \boxed{ \frac{1}{7} + \frac{1}{8} + \frac{1}{18} = \cot^{-1} 3 } \]