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UPSC NDA II 2025 Mathematics Paper

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Question : 50 of 120

Marks: +1, -0

What is the value of

tan[\frac{1}{2}sec^{-1}(\frac{2}{\sqrt{3}})]?

$2-\sqrt{3}$
$2+\sqrt{3}$
$\sqrt{3}-1$
$\sqrt{3}+1$

Solution:

Concept:

Use inverse secant to find the angle, then apply the half-angle formula for tangent.

Explanation:

Let

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

. Then

\sec\theta = \frac{2}{\sqrt{3}}

, so

\cos\theta = \frac{\sqrt{3}}{2}

. This gives

\theta = \frac{\pi}{6}

(or

30^\circ

).

Half the angle:

\frac{1}{2}\theta = \frac{\pi}{12}

(or

15^\circ

). We need

\tan\left(\frac{\pi}{12}\right)

. The known value is

\tan 15^\circ = 2 - \sqrt{3}

.

Answer:

Option A:

2-\sqrt{3}

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