Question

${\tan }^{-1}\left[\frac{1}{4}\right]+{\tan }^{-1}\left[\frac{2}{9}\right]$ equals to

• A. $\frac{1}{2}{\cos }^{-1}\left[\frac{3}{5}\right]$
• B. $\frac{1}{2}{\sin }^{-1}\left[\frac{3}{5}\right]$
• C. $\frac{1}{2}{\tan }^{-1}\left[\frac{3}{5}\right]$
• D. ${\tan }^{-1}\left[\frac{1}{2}\right]$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{1}{2}{\cos }^{-1}\left[\frac{3}{5}\right]$
D ${\tan }^{-1}\left[\frac{1}{2}\right]$

Given, ${\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{2}{36}}\right)$

$={\tan }^{-1}\left(\frac{9+8}{36-2}\right)$

$={\tan }^{-1}\left(\frac{17}{34}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\right)$