Question

If $\theta=\cot^{-1}\,7\cot^{-1}\,8+\cot^{-1}18$ , then $\cot\,\theta$ is equal to

• A. 2
• B. 3
• C. 4
• D. none of these.

Answer 1

Answer: (B)

$\theta = 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}\cdot\frac{1}{8}} + tan^{-1} \frac{1}{18} $
$= tan^{-1} \frac{15}{55} tan^{-1} \frac{1}{18}$
$ = tan^{-1} \frac{3}{11} +tan^{-1} \frac{1}{18} $
$= tan^{-1}\frac{ \frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\cdot\frac{1}{18}} = tan^{-1} \left(\frac{65}{195} \right)$
$ = tan^{-1} \left(\frac{1}{3}\right) = cot^{-1}3$
$\therefore cot\,\theta = 3$