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)

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