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)

$3$

Answer 2

$\theta = tan^{-1} \frac{1}{7}+tan^{-1} \frac{1}{8} + tan^{-1} \frac{1}{18}$ $= tan^{-1} \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\cdot\frac{1}{8}} + tan^{-1} \frac{1}{18}$ $= tan^{-1} \frac{15}{55} +tan^{-1} \frac{1}{18}$ $= tan^{-1} \frac{3}{11}+tan^{-1} \frac{1}{18}$ $= tan^{-1} \frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\cdot\frac{1}{18}}$ $= tan^{-1} \left(\frac{65}{195}\right)$ $= tan^{-1} \left(\frac{1}{3}\right)$ $= cot^{-1}3$ $\therefore cot \,\theta = 3$