If A+ B=-21/65, A+ B=-27/65 and π

Correct Answer is : −3130-{3}{{130}}−1303

Correct Answer is : −3130-{3}{{130}}−1303

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COMEDK 2025 Morning Solved Paper

Question : 149 of 180

Marks: +1, -0

\sin A+\sin B=-\frac{21}{65}, \cos A+\cos B=-\frac{27}{65}

\pi

Solution:

Concept:

Use sum-to-product identities to express

\sin A+\sin B

and

\cos A+\cos B

in terms of

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

and

D=\frac{A-B}{2}

Explanation:

Given:

\sin A+\sin B = -\frac{21}{65}

and

\cos A+\cos B = -\frac{27}{65}

, with

\pi < A-B < 3\pi

Let

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

and

D = \frac{A-B}{2}

Then

\sin A+\sin B = 2\sin S\cos D = -\frac{21}{65}

And

\cos A+\cos B = 2\cos S\cos D = -\frac{27}{65}

Dividing the two equations gives

\tan S = \frac{-21/65}{-27/65} = \frac{21}{27} = \frac{7}{9}

Square and add the equations:

4\cos^2 D (\sin^2 S + \cos^2 S) = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2 = \frac{441+729}{4225} = \frac{1170}{4225} = \frac{18}{65}

Since

\sin^2 S+\cos^2 S = 1

, we have

4\cos^2 D = \frac{18}{65}

, so

\cos^2 D = \frac{18}{260} = \frac{9}{130}

Thus

\cos D = \pm \frac{3}{\sqrt{130}}

Now

\pi < A-B < 3\pi

implies

\frac{\pi}{2} < D < \frac{3\pi}{2}

. In this interval,

\cos D

is negative.

Therefore

\cos D = -\frac{3}{\sqrt{130}}

Answer:

-\frac{3}{\sqrt{130}}

(Option D).

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