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RRB ALP CBT I 25 Nov 2024 Shift 2 Paper

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Question : 3 of 75

Marks: +1, -0

If

4\sin^{2}\theta = 1

\theta

is an acute angle, then the value of

\cos^{2}\theta + \tan^{2}\theta = \_\_\_\_\_\_\_\_

.

$13/12$
$5/13$
$1/3$
$1/4$

Solution:

Concept:

Use the fundamental identity

\sin^{2}\theta + \cos^{2}\theta = 1

and the relation

\tan\theta = \frac{\sin\theta}{\cos\theta}

.

Explanation:

Given

4\sin^{2}\theta = 1

.

So,

\sin^{2}\theta = \frac{1}{4}

.

Since

\theta

is acute,

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

.

Now,

\cos^{2}\theta = 1 - \sin^{2}\theta

.

Thus,

\cos^{2}\theta = 1 - \frac{1}{4} = \frac{3}{4}

.

Also,

\tan^{2}\theta = \frac{\sin^{2}\theta}{\cos^{2}\theta}

.

So,

\tan^{2}\theta = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}

.

Therefore,

\cos^{2}\theta + \tan^{2}\theta = \frac{3}{4} + \frac{1}{3}

.

This gives

\frac{9 + 4}{12} = \frac{13}{12}

.

Answer:

The value is

\frac{13}{12}

, which matches option A.

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