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UPSC CDS II 2025 Mathematics Paper

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Question Numbers: 47-48

Let

\frac{1+sin\theta }{cos\theta }=p+\sqrt{p^{2}+1}

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Question : 47 of 100

Marks: +1, -0

What is sec θ equal to?

p
$\sqrt{p^{2}+1}$
$\frac{1}{\sqrt{p^{2}+1}}$
$\frac{p}{\sqrt{p^{2}+1}}$

Solution:

Concept:

Use the identity

\frac{1+\sin\theta}{\cos\theta} = \sec\theta + \tan\theta

and the relation

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

to solve for

\sec\theta

.

Explanation:

Given

\frac{1+\sin\theta}{\cos\theta} = p + \sqrt{p^{2}+1}

.

We know

\frac{1+\sin\theta}{\cos\theta} = \sec\theta + \tan\theta

. So

\sec\theta + \tan\theta = p + \sqrt{p^{2}+1}

.

Using

\sec^2\theta - \tan^2\theta = (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1

, we get

\sec\theta - \tan\theta = \frac{1}{p + \sqrt{p^{2}+1}}

.

Rationalise:

\frac{1}{p + \sqrt{p^{2}+1}} = \sqrt{p^{2}+1} - p

.

Now add the two equations:

(\sec\theta+\tan\theta) + (\sec\theta-\tan\theta) = (p + \sqrt{p^{2}+1}) + (\sqrt{p^{2}+1} - p)

.

Thus

2\sec\theta = 2\sqrt{p^{2}+1}

. So

\sec\theta = \sqrt{p^{2}+1}

.

Answer:

\sec\theta = \sqrt{p^{2}+1}

, which is option B.

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