If a cosθ – b sinθ = x and a sinθ + b cosθ = y that a2 + b2 =

1.	If a cosθ – b sinθ = x and a sinθ + b cosθ = y that a2 + b2 = x2 + y2.
Answer» Taking RHS =x² + y² Putting the values of x and y, we get (a cos θ – b sin θ)² + (a sin θ + b cos θ)² = a²cos²θ + b² sin²θ – 2ab cos θ sin θ + a²sin²θ + b² cos²θ + 2ab cos θ sin θ = a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ) = a² + b² [∵ cos² θ + sin² θ = 1] = RHS Hence Proved

If a cosθ – b sinθ = x and a sinθ + b cosθ = y that a2 + b2 = x2 + y2.

Answer»

Taking RHS =x² + y²

Putting the values of x and y, we get

(a cos θ – b sin θ)² + (a sin θ + b cos θ)²

= a²cos²θ + b² sin²θ – 2ab cos θ sin θ + a²sin²θ + b² cos²θ + 2ab cos θ sin θ

= a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ)

= a² + b² [∵ cos² θ + sin² θ = 1]

= RHS

Hence Proved