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UPSC NDA I 2026 Mathematics Paper

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Question : 38 of 120

Marks: +1, -0

What is the smallest positive

x

\log _{\sin x} \cos x+\log _{\cos x} \sin x=2

?

$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$

Solution:

Concept:

Use the identity

\log_a b \cdot \log_b a = 1

and set

y = \log_{\sin x}\cos x

.

Explanation:

Let

y = \log_{\sin x}\cos x

. Then

\log_{\cos x}\sin x = \frac{1}{y}

.

The equation becomes

y + \frac{1}{y} = 2

.

Multiplying by

y

:

y^2 + 1 = 2y

, so

(y-1)^2 = 0

.

Thus

y = 1

, meaning

\log_{\sin x}\cos x = 1

.

So

\cos x = \sin x

, giving

\tan x = 1

.

The smallest positive solution is

x = \frac{\pi}{4}

.

Answer:

\frac{\pi}{4}

, which is Option C.

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