Question

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

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

Answer 1

Answer: (C)

$\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$