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UPSC CDS 1 2023 Math Paper

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Question : 52 of 100

Marks: +1, -0

What is the value of

(1+\cot^{2}\theta)(1+\cos\theta)(1-\cos\theta)-(1+\tan^{2}\theta)(1+\sin\theta)(1-\sin\theta)

?

$-1$
0
1
2

Solution:

Concept:

Use trigonometric identities:

1 + \cot^2 \theta = \csc^2 \theta

and

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

, along with

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

.

Explanation:

Start with the given expression:

(1 + \cot^2 \theta)(1 + \cos \theta)(1 - \cos \theta) - (1 + \tan^2 \theta)(1 + \sin \theta)(1 - \sin \theta)

.

Apply identities:

1 + \cot^2 \theta = \csc^2 \theta

and

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

.

Also,

(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta

.

Similarly,

(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta

.

Thus the expression becomes:

\csc^2 \theta \cdot \sin^2 \theta - \sec^2 \theta \cdot \cos^2 \theta

.

Now

\csc^2 \theta \cdot \sin^2 \theta = 1

and

\sec^2 \theta \cdot \cos^2 \theta = 1

.

So the result is

1 - 1 = 0

.

Answer:

0

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