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)

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