Question

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

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

Answer 1

Answer: (C)

Let
${\tan }^{-1}\frac{4}{3}=\theta$
$\Rightarrow \tan \theta =\frac{4}{3}$

$E={\cos }^{-1}\left(\frac{3}{10}\cos \theta +\frac{2}{5}\sin \theta \right)$
$={\cos }^{-1}\left(\frac{3}{10}\times \frac{3}{5}+\frac{2}{5}\cdot \frac{4}{5}\right)$
$={\cos }^{-1}\left(\frac{9}{50}+\frac{8}{25}\right)$
$={\cos }^{-1}\left(\frac{25}{50}\right)$
$={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$