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UPSC CDS 2 2021 Math Paper

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Question : 48 of 100

Marks: +1, -0

What is

\frac{sin^{3}\theta +cos^{3}\theta }{sin\theta +cos\theta }+\frac{sin^{3}\theta -cos^{3}\theta }{sin\theta -cos\theta }

equal to ?

0
1
2
4

Solution:

Concept:

Simplify using algebraic identities for sum and difference of cubes, then apply

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

.

Explanation:

The given expression is:

\frac{\sin^{3}\theta + \cos^{3}\theta}{\sin\theta + \cos\theta} + \frac{\sin^{3}\theta - \cos^{3}\theta}{\sin\theta - \cos\theta}

Use the identities:

a^{3}+b^{3} = (a+b)(a^{2}+b^{2}-ab)

and

a^{3}-b^{3} = (a-b)(a^{2}+b^{2}+ab)

.

Apply them to each fraction:

\frac{(\sin\theta+\cos\theta)(\sin^{2}\theta+\cos^{2}\theta - \sin\theta\cos\theta)}{\sin\theta+\cos\theta} + \frac{(\sin\theta-\cos\theta)(\sin^{2}\theta+\cos^{2}\theta + \sin\theta\cos\theta)}{\sin\theta-\cos\theta}

Cancel the common factors:

(\sin^{2}\theta+\cos^{2}\theta - \sin\theta\cos\theta) + (\sin^{2}\theta+\cos^{2}\theta + \sin\theta\cos\theta)

Combine like terms:

(\sin^{2}\theta+\cos^{2}\theta) + (\sin^{2}\theta+\cos^{2}\theta) + (-\sin\theta\cos\theta + \sin\theta\cos\theta)

The terms with

\sin\theta\cos\theta

cancel.

\sin^{2}\theta+\cos^{2}\theta = 1

. So we get:

1 + 1 = 2

.

Answer:

2

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