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UPSC CDS 2 2020 Math Paper

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Question : 47 of 100

Marks: +1, -0

If

\cos heta + \sec heta = k

, then what is the value of

\sin^{2} heta - an^{2} heta

?

4 - k
$4 - k^2$
$k^2 - 4$
$k^2 + 2$

Solution:

Concept:

Express

\sin^2 heta - an^2 heta

in terms of

\cos^2 heta

and

\sec^2 heta

, then use the given relation.

Explanation:

Given

\cos heta + \sec heta = k

.

Recall

\sec\theta = \frac{1}{\cos\theta}

, so

\cos\theta + \frac{1}{\cos\theta} = k

.

Square both sides:

\left(\cos \theta + \frac{1}{\cos \theta}\right)^2 = k^2

.

Expand:

\cos^2\theta + \frac{1}{\cos^2\theta} + 2 = k^2

.

Thus

\cos^2 heta + \sec^2 heta = k^2 - 2

.

Now compute

\sin^2 heta - an^2 heta

.

\sin^2 heta = 1 - \cos^2 heta

and

an^2 heta = \sec^2 heta - 1

.

So

\sin^2 heta - an^2 heta = (1 - \cos^2 heta) - (\sec^2 heta - 1)

.

Simplify:

1 - \cos^2 heta - \sec^2 heta + 1 = 2 - (\cos^2 heta + \sec^2 heta)

.

Substitute

\cos^2 heta + \sec^2 heta = k^2 - 2

.

Therefore,

2 - (k^2 - 2) = 4 - k^2

.

Answer:

4 - k^2

(Option B).

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