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 :

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Inverse Trigonometric Functions

If the inverse trigonometric functions take principal values, then

cos^{- 1} (\frac{3}{10} cos (tan^{- 1} (\frac{4}{3})) + \frac{2}{5} sin (tan^{- 1} (\frac{4}{3})))

is equal to :

Answer: (C)

Let

θ = tan^{- 1} \frac{4}{3}

, so

tan θ = \frac{4}{3}

. Then

cos θ = \frac{3}{5}

and

sin θ = \frac{4}{5}

.

Substitute:

\frac{3}{10} \cdot \frac{3}{5} + \frac{2}{5} \cdot \frac{4}{5} = \frac{9}{50} + \frac{8}{25} = \frac{9}{50} + \frac{16}{50} = \frac{25}{50} = \frac{1}{2}

.

Therefore

cos^{- 1} (\frac{1}{2}) = \frac{π}{3}

.

Answer: (C)

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