Question

Prove that cot⁻¹7 + cot⁻¹8 + cot⁻¹18 = cot⁻¹3

Answer 1

${cot}^{ - 1} 7 + {cot}^{ - 1} 8 + {cot}^{ - 1} 18$

$= {tan}^{ - 1} \frac{1}{7} + {tan}^{ - 1} \frac{1}{8} + {tan}^{ - 1} \frac{1}{18}$

$= {tan}^{ - 1} \frac{ \frac{1}{7} + \frac{1}{8} }{1 - \frac{1}{7} \times \frac{1}{8} } + {tan}^{ - 1} \frac{1}{18}$

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

$= {tan}^{ - 1} \frac{ \frac{3}{11} + \frac{1}{18} }{1 - \frac{3}{11} \times \frac{1}{18} }$

$= {tan}^{ - 1} \frac{65}{195}$

$= {tan}^{ - 1} \frac{1}{3}$

$= {cot}^{ - 1} 3$

( Proved)

We know,

${tan}^{ - 1}a +{tan}^{ - 1}b = {tan}^{ - 1} \frac{a + b}{1 - ab}$