question_answer

                                             \[\] Prove that \[{{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\cot }^{-1}}3.\]

Answer:

\[LHS={{\cot }^{-\,1}}7+{{\cot }^{-\,1}}8+{{\cot }^{-\,1}}18\] \[={{\tan }^{-\,1}}\frac{1}{7}+{{\tan }^{-\,1}}\frac{1}{8}+{{\tan }^{-\,1}}\frac{1}{18}\] \[\left[ \because {{\cot }^{-\,1}}x={{\tan }^{-\,1}}\left( \frac{1}{x} \right) \right]\] \[={{\tan }^{-\,1}}\left( \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\times \frac{1}{8}} \right)+{{\tan }^{-\,1}}\frac{1}{18}\] \[={{\tan }^{-\,1}}\left( \frac{\frac{15}{56}}{\frac{56-1}{56}} \right)+{{\tan }^{-\,1}}\frac{1}{18}\] \[={{\tan }^{-\,1}}\left( \frac{\frac{15}{56}}{\frac{55}{56}} \right)+{{\tan }^{-\,1}}\frac{1}{18}\] \[={{\tan }^{-\,1}}\left( \frac{3}{11} \right)+{{\tan }^{-\,1}}\frac{1}{18}\] \[={{\tan }^{-\,1}}\left( \frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\times 1\frac{1}{18}} \right)+{{\tan }^{-\,1}}\left( \frac{\frac{54+11}{11\times 18}}{\frac{198-3}{11\times 18}} \right)\] \[={{\tan }^{-\,1}}\left( \frac{65}{195} \right)+{{\tan }^{-\,1}}\left( \frac{1}{3} \right)={{\cot }^{-\,1}}3\] \[=RHS\]                                  Hence proved.