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UPSC NDA II 2018 Mathematics Solved Paper

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Question : 36 of 120

Marks: +1, -0

What is

\frac{2tan\theta }{1+tan^{2}\theta }

equal to?

$cos 2θ$
$tan 2θ$
$sin 2θ$
$cosec 2θ$

Solution:

Concept:

Use trigonometric identities:

1+\tan^2\theta = \sec^2\theta

and

\sin 2\theta = 2\sin\theta\cos\theta

.

Explanation:

Start with

\frac{2\tan\theta}{1+\tan^2\theta}

.

Replace denominator:

1+\tan^2\theta = \sec^2\theta

.

So expression becomes

\frac{2\tan\theta}{\sec^2\theta}

.

Write

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

and

\sec^2\theta = \frac{1}{\cos^2\theta}

.

Thus

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

.

And

2\sin\theta\cos\theta = \sin 2\theta

.

Answer:

\sin 2\theta

, which corresponds to option C.

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