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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 : 36 of 120

Marks: +1, -0

What is cos (α – β) equal to?

Solution:

Concept:

Use the identity

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

and the cosine difference formula

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

. (Take positive roots as angles are acute).

Write

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

Apply the cosine difference formula:

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

Substitute the values:

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

Thus,

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

Answer:

Option A:

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

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