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Question

Mathematics Question on Inverse Trigonometric Functions

The period of tan3θ\tan 3\theta is

A

π\pi

B

3π4\frac{3 \pi}{4}

C

π2\frac{\pi}{2}

D

None of these

Answer

None of these

Explanation

Solution

We need to find period of tanθ\tan \theta
tanθ\tan \theta is of period π\pi, so that tan3θ\tan 3 \theta is of period π3\frac{\pi}{3}.