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Question

Mathematics Question on Inverse Trigonometric Functions

The value of tan⁡−1(√3)−sec⁡−1(2√3)\tan^{-1}(√3)-\sec^{-1}(\dfrac{2}{√3}) is ?

A

2π3\dfrac{2π}{3}

B

π4\dfrac{π}{4}

C

π3\dfrac{π}{3}

D

π2\dfrac{π}{2}

E

π6\dfrac{π}{6}

Answer

π2\dfrac{π}{2}

Explanation

Solution

tan⁡−1(√3)−sec⁡−1(2√3)\tan^{-1}(√3)-\sec^{-1}(\dfrac{2}{√3})
=π3+π6=\dfrac{\pi}{3}+\dfrac{\pi}{6}

=π2=\dfrac{\pi}{2}
So, the correct option is (D) :π2\dfrac{π}{2}