Test Index

UPSC CDS II 2025 Mathematics Paper

Show Para Hide Para

Question Numbers: 47-48

Let

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

© examsnet.com

Question : 48 of 100

Marks: +1, -0

What is tan θ equal to?

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

Solution:

Concept:

Use the identities

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

and

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

.

Explanation:

We are given

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

.

This means

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

.

Using the identity,

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

.

Rationalising the denominator gives

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

.

Subtract the second equation from the first:

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

.

Simplifying,

2\tan\theta = 2p

.

Thus

\tan\theta = p

.

Answer:

p

(Option A)

© examsnet.com

Go to Question:

Prev Question

Next Question