If the inverse trigonometric functions take p | vedclass

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Question

(C) Let $	an ^{-1} \left(\frac{4}{3}ight) = 	heta$,which implies $	an 	heta = \frac{4}{3}$.From the right-angled triangle with opposite side $4$

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Let $	an ^{-1} \left(\frac{4}{3}ight) = 	heta$,which implies $	an 	heta = \frac{4}{3}$.From the right-angled triangle with opposite side $4$ and adjacent side $3$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.Thus,$\cos 	heta = \frac{3}{5}$ and $\sin 	heta = \frac{4}{5}$.Let the given expression be $E = \cos ^{-1}\left(\frac{3}{10} \cos 	heta + \frac{2}{5} \sin 	hetaight)$.Substituting the values of $\cos 	heta$ and $\sin 	heta$:$E = \cos ^{-1}\left(\frac{3}{10} 	imes \frac{3}{5} + \frac{2}{5} 	imes \frac{4}{5}ight)$$E = \cos ^{-1}\left(\frac{9}{50} + \frac{8}{25}ight)$$E = \cos ^{-1}\left(\frac{9}{50} + \frac{16}{50}ight)$$E = \cos ^{-1}\left(\frac{25}{50}ight)$$E = \cos ^{-1}\left(\frac{1}{2}ight)$Since the principal value of $\cos ^{-1}\left(\frac{1}{2}ight)$ is $\frac{\pi}{3}$,the final answer is $\frac{\pi}{3}$.

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

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