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UPSC CDS 2 2022 Math Paper

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Question : 59 of 100

Marks: +1, -0

What is the minimum value of

3600 \sec^{2}\theta + 121 \csc^{2}\theta

0 < \theta < \frac{\pi}{2}

?

1320
2401
3721
5041

Solution:

Concept:

For

0 < \theta < \frac{\pi}{2}

, the expression

a \sec^2\theta + b \csc^2\theta

attains its minimum value

(\sqrt{a} + \sqrt{b})^2

.

Explanation:

Given expression:

3600 \sec^2\theta + 121 \csc^2\theta

.

Identify

a = 3600

and

b = 121

.

Compute square roots:

\sqrt{3600} = 60

,

\sqrt{121} = 11

.

Their sum:

60 + 11 = 71

.

Square the sum:

71^2 = 5041

.

Thus, the minimum value is

5041

.

Answer:

5041

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