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prove that

cot^-1 (7) + cot^-1 (8) + cot^-1 (18) = cot^-1 (3)

$\cot^{- 1} (7) + \cot^{- 1} (8) + \cot^{- 1} (18) = \tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{8}) + \tan^{- 1} (\frac{1}{18}) = [\tan^{- 1} \frac{1}{7} + \tan^{- 1} (\frac{1}{8})] + \tan^{- 1} (\frac{1}{18}) = \tan^{- 1} \frac{\frac{1}{7} + \frac{1}{8}}{1 - (\frac{1}{7}) (\frac{1}{8})} + \tan^{- 1} \frac{1}{18} (u \sin g \tan^{- 1} A + \tan^{- 1} B = \tan^{- 1} (\frac{A + B}{1 - A B})) = \tan^{- 1} \frac{8 + 7}{56 - 1} + \tan^{- 1} \frac{1}{18} = \tan^{- 1} \frac{3}{11} + \tan^{- 1} \frac{1}{18} = \tan^{- 1} \frac{\frac{3}{11} + \frac{1}{18}}{1 - (\frac{3}{11}) (\frac{1}{18})} = \tan^{- 1} \frac{54 + 11}{198 - 3} = \tan^{- 1} \frac{65}{195} = \tan^{- 1} \frac{1}{3} = \cot^{- 1} (3) = R . H . S H e n c e p r o v e d .$

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