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Question

Mathematics Question on Inverse Trigonometric Functions

If the inverse trigonometric functions take principal values, then
cos−1(310cos(tan−1(43))+25sin(tan−1(43)))cos^{-1} ( \frac{3}{10} cos (tan^{-1} (\frac{4}{3})) + \frac{2}{5} sin (tan^{-1} (\frac{4}{3})))
is equal to :

A

0

B

π4\frac{\pi}{4}

C

π3\frac{\pi}{3}

D

π6\frac{\pi}{6}

Answer

π3\frac{\pi}{3}

Explanation

Solution

The correct answer is (C) : π3\frac{\pi}{3}
cos−1(310cos(tan−1(43))+25sin(tan−1(43)))cos^{-1} ( \frac{3}{10} cos (tan^{-1} (\frac{4}{3})) + \frac{2}{5} sin (tan^{-1} (\frac{4}{3})))

=cos−1(310.35+25.45)= cos^{-1} ( \frac{3}{10} . \frac{3}{5} + \frac{2}{5} . \frac{4}{5} )

=cos−1(12)=π3= cos^{-1} (\frac{1}{2}) = \frac{π}{3}