Question

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

Answer 1

LHS : = ${\cot }^{-1}\left(7\right)+{\cot }^{-1}\left(8\right)+{\cot }^{-1}\left(18\right)$

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

$(\because {\cot }^{-1}(x)={\tan }^{-1}\left(\frac{1}{x}\right))$

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

$(\because {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right))$

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

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

$={\tan }^{-1}\left(\frac{54+11}{198-3}\right)\cdot {\tan }^{-1}\left(\frac{65}{195}\right)$

$={\tan }^{-1}\left(\frac{1}{3}\right)$

${\cot }^{-1}\left(3\right)$