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UPSC CDS 1 2020 Math Paper

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Question : 64 of 100

Marks: +1, -0

What is the value of

\frac{{1 - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }} + 4

equal to?

0
1
2
5

Solution:

Concept:

Use the identity

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

and

(a+b)^2 - 2ab = a^2 + b^2

to simplify the denominator.

Explanation:

Start with the expression:

\frac{1 - 2\sin^2\theta \cos^2\theta}{\sin^4\theta + \cos^4\theta} + 4

.

Rewrite

\sin^4\theta + \cos^4\theta

using

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

.

Since

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

, the denominator becomes

1 - 2\sin^2\theta\cos^2\theta

.

So the fraction becomes

\frac{1 - 2\sin^2\theta\cos^2\theta}{1 - 2\sin^2\theta\cos^2\theta} = 1

.

Finally, add 4:

1 + 4 = 5

.

Answer:

5

(Option D)

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