Question

If the inverse trigonometric functions take principal values, then ${\cos }^{-1}(\frac{3}{10}\cos \left({\tan }^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}\sin \left({\tan }^{-1}\left(\frac{4}{3}\right)\right))$ is equal to:

• A. $\frac{\pi }{3}$
• B. $0$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{3}$

${\cos }^{-1}(\frac{3}{10}\cos \left({\tan }^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}\sin \left({\tan }^{-1}\left(\frac{4}{3}\right)\right))$
$={\cos }^{-1}(\frac{3}{10}\cdot \frac{3}{5}+\frac{2}{5}\cdot \frac{4}{5})$ $={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$