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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 32 of 120

Marks: +1, -0

Let

4\sin^{2}x = 3

0 \le x \le \pi

\tan 3x

equal to?

$-2$
$-1$
0
1

Solution:

Concept:

Solving a trigonometric equation to find

x

and then evaluating

\tan 3x

.

Explanation:

We are given

4\sin^{2}x = 3

for

0 \le x \le \pi

.

Divide both sides by

4

:

\sin^{2}x = \frac{3}{4}

.

Take the square root:

\sin x = \frac{\sqrt{3}}{2}

(positive because

\sin x \ge 0

in

[0,\pi]

).

We know

\sin 60^\circ = \frac{\sqrt{3}}{2}

and

\sin(180^\circ - 60^\circ) = \sin 120^\circ = \frac{\sqrt{3}}{2}

.

Thus

x = 60^\circ

or

x = 120^\circ

within the given interval.

Now compute

\tan 3x

for each possible

x

:

If

x = 60^\circ

, then

3x = 180^\circ

and

\tan 180^\circ = 0

.

If

x = 120^\circ

, then

3x = 360^\circ

and

\tan 360^\circ = 0

.

Both cases give the same value.

Answer:

0

(Option C).

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