Question

$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$ is equal to:

• A. 1
• B. 2
• C. $\frac{1}{4}$
• D. $\frac{5}{4}$

Answer 1

Answer: (B)

$\tan \left(2\left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right) $
$ =\tan \left[2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right] $
$ =2$