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UPSC NDA II 2025 Mathematics Paper

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Question Numbers: 27-28

For the following two (02) items:

Let

p = cosec 20

° and

q = cosec 70

°.

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Question : 28 of 120

Marks: +1, -0

What is

\frac{p^{2}+q^{2}}{p^{2}q^{2}}

equal to?

$1/2$
1
$3/2$
2

Solution:

Concept:

Use the relation

p = \csc 20^\circ

,

q = \csc 70^\circ

and the identity

\sin 70^\circ = \cos 20^\circ

to simplify the expression exactly.

Explanation:

Given

p = \csc 20^\circ = \frac{1}{\sin 20^\circ}

and

q = \csc 70^\circ = \frac{1}{\sin 70^\circ}

.

Since

\sin 70^\circ = \cos 20^\circ

, we have

q = \frac{1}{\cos 20^\circ}

.

Then

p^2 = \frac{1}{\sin^2 20^\circ}

and

q^2 = \frac{1}{\cos^2 20^\circ}

.

Now

p^2 + q^2 = \frac{1}{\sin^2 20^\circ} + \frac{1}{\cos^2 20^\circ} = \frac{\cos^2 20^\circ + \sin^2 20^\circ}{\sin^2 20^\circ \cos^2 20^\circ} = \frac{1}{\sin^2 20^\circ \cos^2 20^\circ}

.

Also

p^2 q^2 = \frac{1}{\sin^2 20^\circ} \cdot \frac{1}{\cos^2 20^\circ} = \frac{1}{\sin^2 20^\circ \cos^2 20^\circ}

.

Hence

\frac{p^2 + q^2}{p^2 q^2} = \frac{ \frac{1}{\sin^2 20^\circ \cos^2 20^\circ} }{ \frac{1}{\sin^2 20^\circ \cos^2 20^\circ} } = 1

.

Answer:

B. 1

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