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UPSC NDA I 2012 Mathematics Paper

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For the next Three (3) questions that follow:

Let

\sin\; (A+B)=1

and

\sin\; (A-B)=\frac{1}{2}

, where

A, B

\in \left[0, \;\frac{\pi}{2}\right]

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Question : 27 of 120

Marks: +1, -0

What is

\sin^2 A - \sin^2 B

0
$\frac{1}{2}$
1
2

Solution:

\sin^2 60^{\circ} - \sin^2 30^{\circ}

\frac{3}{4} - \frac{1}{4} = \frac{1}{2}

.

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