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COMEDK 2025 Afternoon Solved Paper

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Question : 172 of 180

Marks: +1, -0

The value of

\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)

is

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{5 \pi}{6}$
$\frac{7 \pi}{6}$

Solution:

Concept:

The inverse trigonometric function

\tan^{-1}

returns a principal value restricted to the interval

(-\frac{\pi}{2}, \frac{\pi}{2})

.

Explanation:

First, find

\tan\left(\frac{7\pi}{6}\right)

.

\frac{7\pi}{6} = \pi + \frac{\pi}{6}

, so

\tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}

.

The principal range of

\tan^{-1}x

is

-\frac{\pi}{2} < y < \frac{\pi}{2}

.

Since

\frac{1}{\sqrt{3}}

lies within this range, we compute

y = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}

.

Answer:

\frac{\pi}{6}

(Option A)

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