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

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