The value of 3 ^{-1}≤ft(1/2)=

Correct Answer is : tan⁡−1(112) ^{-1}≤ft(11/2)tan−1(211)

Correct Answer is : tan⁡−1(112) ^{-1}≤ft(11/2)tan−1(211)

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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 117 of 150

Marks: +1, -0

The value of

3 \tan ^{-1}\left(\frac{1}{2}\right)=\ldots

Solution:

Concept:

Use the triple-angle formula for tangent and then apply the principal value range of inverse tangent.

Explanation:

Let

\theta = \tan^{-1}\left(\frac{1}{2}\right)

Therefore,

\tan \theta = \frac{1}{2}

We need to find

3\theta

Use the formula:

\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}

Substitute

\tan\theta = \frac{1}{2}

\tan 3\theta = \frac{3\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)^3}{1 - 3\left(\frac{1}{2}\right)^2}

= \frac{\frac{3}{2} - \frac{1}{8}}{1 - \frac{3}{4}} = \frac{\frac{12}{8} - \frac{1}{8}}{\frac{1}{4}}

= \frac{\frac{11}{8}}{\frac{1}{4}} = \frac{11}{2}

So,

\tan 3\theta = \frac{11}{2}

Since

0 < \theta < \frac{\pi}{6}

, we get

0 < 3\theta < \frac{\pi}{2}

Thus,

3\theta

lies in the principal range of

\tan^{-1}

Hence,

3\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{11}{2}\right)

Answer:

\tan^{-1}\left(\frac{11}{2}\right)

Correct option: D

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