Question

Prove that:
cot⁻¹ 7 + cot​⁻¹ 8 + cot​⁻¹ 18 = cot​⁻¹ 3 .

Answer 1

Solving L.H.S, we get:

\[\cot^{- 1} 7 + \cot^{- 1} 8 + \cot^{- 1} 18 = \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8} + \tan^{- 1} \frac{1}{18}\]

\[\left\{ \text { Using, }\tan^{- 1} A + \tan^{- 1} B = \tan^{- 1} \left( \frac{A + B}{1 - AB} \right) \right\}\]

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

\[ = \tan^{- 1} \left( \frac{15}{56 - 1} \right) + \tan^{- 1} \frac{1}{18}\]

\[ = \tan^{- 1} \frac{3}{11} + \tan^{- 1} \frac{1}{18}\]

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

\[ = \tan^{- 1} \left( \frac{54 + 11}{198 - 3} \right)\]

\[ = \tan^{- 1} \left( \frac{65}{195} \right)\]

\[ = \tan^{- 1} \frac{1}{3}\]

\[ = \cot^{- 1} \left( 3 \right) = RHS\]

Hence proved.