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UPSC CDS 1 2022 Math Paper

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Question : 55 of 100

Marks: +1, -0

What is

(\sec^2\alpha + \tan\alpha \cdot \tan\beta - \tan^2\alpha)^2 + (\tan\alpha - \tan\beta)^2 - \sec^2\alpha \cdot \sec^2\beta

equal to?

$-1$
0
1
2

Solution:

Concept:

Use the identity

\sec^2\theta - \tan^2\theta = 1

to simplify the expression.

Explanation:

Start with the given expression:

(\sec^2\alpha + \tan\alpha \tan\beta - \tan^2\alpha)^2 + (\tan\alpha - \tan\beta)^2 - \sec^2\alpha \cdot \sec^2\beta

Rewrite

\sec^2\alpha - \tan^2\alpha

as

1

.

The first term becomes

(1 + \tan\alpha \tan\beta)^2

.

Also write

\sec^2\alpha = \tan^2\alpha + 1

and

\sec^2\beta = \tan^2\beta + 1

.

Thus the expression is:

(1 + \tan\alpha \tan\beta)^2 + (\tan\alpha - \tan\beta)^2 - (\tan^2\alpha + 1)(\tan^2\beta + 1)

Expand each part:

(1 + \tan\alpha \tan\beta)^2 = 1 + 2\tan\alpha\tan\beta + \tan^2\alpha \tan^2\beta

(\tan\alpha - \tan\beta)^2 = \tan^2\alpha + \tan^2\beta - 2\tan\alpha\tan\beta

Add them:

1 + 2\tan\alpha\tan\beta + \tan^2\alpha\tan^2\beta + \tan^2\alpha + \tan^2\beta - 2\tan\alpha\tan\beta

Simplify:

1 + \tan^2\alpha\tan^2\beta + \tan^2\alpha + \tan^2\beta

Now expand

(\tan^2\alpha+1)(\tan^2\beta+1) = \tan^2\alpha\tan^2\beta + \tan^2\alpha + \tan^2\beta + 1

Subtract this from the previous sum:

[1 + \tan^2\alpha\tan^2\beta + \tan^2\alpha + \tan^2\beta] - [\tan^2\alpha\tan^2\beta + \tan^2\alpha + \tan^2\beta + 1] = 0

All terms cancel, leaving zero.

Answer:

0

(Option B)

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