If x θ + y θ = z, then what is the value of (x θ - y θ)²?

Correct Answer is : x2+y2−z2x² + y² - z²x2+y2−z2

Correct Answer is : x2+y2−z2x² + y² - z²x2+y2−z2

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UPSC NDA 2 2016 Mathematics Solved Paper

Question : 45 of 120

Marks: +1, -0

x \cos \theta + y \sin \theta = z

(x \sin \theta - y \cos \theta)^2

Solution:

Concept:

Use the Pythagorean identity

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

to simplify the expression.

Explanation:

Given:

x \cos \theta + y \sin \theta = z

Square both sides:

x^2 \cos^2 \theta + y^2 \sin^2 \theta + 2xy \sin \theta \cos \theta = z^2

So,

2xy \sin \theta \cos \theta = z^2 - x^2 \cos^2 \theta - y^2 \sin^2 \theta

Now compute

(x \sin \theta - y \cos \theta)^2

Expand:

x^2 \sin^2 \theta + y^2 \cos^2 \theta - 2xy \sin \theta \cos \theta

Substitute the expression for

2xy \sin \theta \cos \theta

= x^2 \sin^2 \theta + y^2 \cos^2 \theta - (z^2 - x^2 \cos^2 \theta - y^2 \sin^2 \theta)

Simplify:

= x^2 \sin^2 \theta + y^2 \cos^2 \theta - z^2 + x^2 \cos^2 \theta + y^2 \sin^2 \theta

Group:

= x^2(\sin^2 \theta + \cos^2 \theta) + y^2(\sin^2 \theta + \cos^2 \theta) - z^2

Using identity:

= x^2 + y^2 - z^2

Answer:

Option A:

x^2 + y^2 - z^2

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