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UPSC NDA I 2019 Mathematics Solved Paper

Question Numbers: 36-37

Direction: Consider the following for the next 02 (Two) items:

It is given that cos (θ – α) = a, cos (θ – β) = b.

Question : 37 of 120

Marks: +1, -0

What is sin2 (α – β) + 2ab cos (α – β) equal to?

Solution:

Concept:

Use the identities

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

\cos(A-B) = \cos A \cos B + \sin A \sin B

, and

\sin(A-B) = \sin A \cos B - \cos A \sin B

Explanation:

Given

\cos(\theta - \alpha) = a

and

\cos(\theta - \beta) = b

From

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

, we get

\sin(\theta - \alpha) = \sqrt{1 - a^2}

and

\sin(\theta - \beta) = \sqrt{1 - b^2}

From a previous step,

\cos(\alpha - \beta) = ab + \sqrt{1 - a^2} \sqrt{1 - b^2}

Now express

\sin(\alpha - \beta) = \sin[(\theta - \beta) - (\theta - \alpha)]

Using the sine difference formula:

\sin(\alpha - \beta) = \sin(\theta - \beta)\cos(\theta - \alpha) - \cos(\theta - \beta)\sin(\theta - \alpha)

Substitute the given values:

\sin(\alpha - \beta) = a\sqrt{1 - b^2} - b\sqrt{1 - a^2}

Now compute the required expression:

\sin^2(\alpha - \beta) + 2ab\cos(\alpha - \beta)

Substitute

\sin(\alpha - \beta)

and

\cos(\alpha - \beta)

: it becomes

(a\sqrt{1 - b^2} - b\sqrt{1 - a^2})^2 + 2ab[ab + \sqrt{1 - a^2}\sqrt{1 - b^2}]

Expand:

a^2(1 - b^2) + b^2(1 - a^2) - 2ab\sqrt{1 - a^2}\sqrt{1 - b^2} + 2a^2b^2 + 2ab\sqrt{1 - a^2}\sqrt{1 - b^2}

Cancel the terms

-2ab\sqrt{1 - a^2}\sqrt{1 - b^2}

and

+2ab\sqrt{1 - a^2}\sqrt{1 - b^2}

Simplify:

a^2 - a^2b^2 + b^2 - a^2b^2 + 2a^2b^2 = a^2 + b^2

Answer:

a^2 + b^2

(Option A).

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