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UPSC NDA I 2026 Mathematics Paper

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Question : 33 of 120

Marks: +1, -0

In a triangle

A B C, \sin A=\cos B+\cos C

\tan \left(\frac{B}{2}\right)+\cot \left(\frac{B}{2}\right)

equal to ?

1
$\sqrt{2}$
$\sqrt{3}$
2

Solution:

Concept:

Use

\sin A=\sin(B+C)

and sum-to-product identities.

Explanation:

Since

A+B+C=\pi

, we have

\sin A=\sin(B+C)

.

Given

\sin(B+C)=\cos B+\cos C

,

2\sin\frac{B+C}{2}\cos\frac{B+C}{2} = 2\cos\frac{B+C}{2}\cos\frac{B-C}{2}.

Cancelling

2\cos\frac{B+C}{2}\neq0

,

\sin\frac{B+C}{2} = \cos\frac{B-C}{2}.

Thus the intended fixed case gives

B=90^\circ

.

So,

\tan\frac{B}{2}+\cot\frac{B}{2} = \tan45^\circ+\cot45^\circ = 1+1=2.

Answer:

D. 2

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