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UPSC CDS 2 2021 Math Paper

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Question : 55 of 100

Marks: +1, -0

If

x = p\sin A\cos B

y = p\sin A\sin B

z = p\cos A

, then what is the value of

x^{2} + y^{2} + z^{2}

?

$-p^{2}$
0
$p^{2}$
$2p^{2}$

Solution:

Concept:

Use the trigonometric identity

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

to simplify the sum of squares.

Explanation:

Given:

x = p \sin A \cos B

,

y = p \sin A \sin B

,

z = p \cos A

.

Compute each square:

x^2 = p^2 \sin^2 A \cos^2 B

,

y^2 = p^2 \sin^2 A \sin^2 B

,

z^2 = p^2 \cos^2 A

.

Add them:

x^2 + y^2 + z^2 = p^2(\sin^2 A \cos^2 B + \sin^2 A \sin^2 B + \cos^2 A)

.

Factor

\sin^2 A

from the first two terms:

= p^2[\sin^2 A(\cos^2 B + \sin^2 B) + \cos^2 A]

.

Using

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

:

= p^2[\sin^2 A(1) + \cos^2 A] = p^2(\sin^2 A + \cos^2 A)

.

Again using

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

:

= p^2 \times 1 = p^2

.

Answer:

p^2

(Option C).

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