If a = xcosθ + ysinθ and b = xsinθ − ycosθ, then a² + b² is equal to:

Correct — A. Square each expression and add.

a² = (x cosθ + y sinθ)² = x²cos²θ + 2xy sinθ cosθ + y²sin²θ
b² = (x sinθ − y cosθ)² = x²sin²θ − 2xy sinθ cosθ + y²cos²θ

The cross terms are equal and opposite, so they cancel:
a² + b² = x²(cos²θ + sin²θ) + y²(sin²θ + cos²θ)

Since cos²θ + sin²θ = 1, this collapses to x² + y², the expression on option (a).

Why the others are wrong

• (b)
  Option (b) shows x + y, which is degree 1 while a² + b² is degree 2. Double x and y and a² + b² quadruples while x + y only doubles, so the two cannot be the same thing.
• (c)
  Option (c) shows x² − y², the shape you expect if the minus sign in b carried through. It does not: squaring −y cosθ gives +y²cos²θ, so both squares arrive positive.
• (d)
  Option (d) shows x − y, degree 1 again, and it turns negative whenever y exceeds x. A sum of two squares never does.

This is the Pythagorean identity sin²θ + cos²θ = 1 hidden inside an algebraic expansion.

Mechanically: expand both squares, watch the 2xy cross terms cancel because one carries a plus and the other a minus, group the x² and y² terms, then apply the identity twice.

Structurally: the map (x, y) → (a, b) here is orthogonal, and orthogonal maps leave x² + y² alone. That is why the answer is free of θ.

Key facts

• sin²θ + cos²θ = 1 for every value of θ.
• If a = x cosθ + y sinθ and b = x sinθ − y cosθ then a² + b² = x² + y², whatever θ is.
• Putting θ = 0 gives a = x and b = −y, so a² + b² = x² + y² — a fast sanity check.
• (p + q)² + (p − q)² = 2p² + 2q² is the same cross-term cancellation in plain algebra.

Common traps

• Expanding one square carefully and assuming the second behaves identically
• Letting the minus sign inside b survive into the answer as x² − y²
• Testing one convenient value of θ and treating the coincidence as proof without the algebra

No directly related past PYQ was found.