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UPSC NDA Math Model Paper 4

Comprehension:(Que No. 78 - 80)

Direction: Let sin (A + B) = 1 and sin (A - B) = 1 / 2, where

A, B \in \left[0, \frac{\pi}{2}\right].

Question : 78 of 120

Marks: +1, -0

What is

\sin^{2} A - \sin^{2} B

equal to?

Solution:

Given that

\sin(A+B) = 1

and

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

In the given domain the function

\sin x

is one one.

Therefore,

A+B = \frac{\pi}{2}

and

A-B = \frac{\pi}{6}

Solving the both equations for A we get,

2A = \frac{4\pi}{6}

A = \frac{\pi}{3}

Therefore,

A = \frac{\pi}{3}

Substituting in the value of A+ B we get that

B = \frac{\pi}{6}

Therefore,

\sin^{2} A - \cos^{2} B = \sin^{2}\left(\frac{\pi}{3}\right) - \sin^{2}\left(\frac{\pi}{6}\right)

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

= \frac{1}{2}

Therefore,

\sin^{2} A - \sin^{2} B = \frac{1}{2}

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