Question

${\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)$ is equal to

• A. $\frac{1}{2}{\cos }^{-1}\left(\frac{3}{5}\right)$
• B. $\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• D. ${\cos }^{-1}\left(\frac{8}{9}\right)$

Answer 1

Answer: (C)

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

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

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

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

$={\tan }^{-1}\left(\frac{\frac{9+8}{36}}{\frac{34}{36}}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\right)$ or ${\sin }^{-1}\frac{1}{\sqrt{5}}$ or ${\cos }^{-1}\frac{2}{\sqrt{5}}$