If a cos θ + b sin θ and a sin θ − b cos θ = 3, then a2 + b2 =

1.	If a cos θ + b sin θ and a sin θ − b cos θ = 3, then a2 + b2 = A. 7 B. 12 C. 25 D. None of these
Answer» Given: a cosθ + b sinθ = 4 Squaring both sides, we get (a cosθ + b sinθ)² = 4² ⇒ a² cos²θ + b² sin²θ + 2ab sin θ cos θ = 16 …(i) and a sinθ – b cosθ = 3 Squaring both sides, we get (a sinθ – b cosθ)² = 3² ⇒ a² sin²θ + b² cos²θ – 2ab sinθ cosθ = 9 …(ii) To find: a²+ b² Adding (i) and (ii), we get a² cos²θ + b² sin²θ + 2ab sinθ cosθ + a² sin²θ + b²cos²θ – 2ab sinθ cosθ = 16 + 9 ⇒ a² (sin²θ + cos²θ) + b² (sin² θ + cos²θ) = 25 ⇒ a² + b² = 25 [∵ sin²θ + cos²θ = 1]

If a cos θ + b sin θ and a sin θ − b cos θ = 3, then a2 + b2 = A. 7 B. 12 C. 25 D. None of these

Answer»

Given: a cosθ + b sinθ = 4

Squaring both sides, we get

(a cosθ + b sinθ)² = 4²

⇒ a² cos²θ + b² sin²θ + 2ab sin θ cos θ = 16 …(i)

and a sinθ – b cosθ = 3

Squaring both sides, we get

(a sinθ – b cosθ)² = 3²

⇒ a² sin²θ + b² cos²θ – 2ab sinθ cosθ = 9 …(ii)

To find: a²+ b²

Adding (i) and (ii), we get

a² cos²θ + b² sin²θ + 2ab sinθ cosθ + a² sin²θ + b²cos²θ – 2ab sinθ cosθ = 16 + 9

⇒ a² (sin²θ + cos²θ) + b² (sin² θ + cos²θ) = 25

⇒ a² + b² = 25 [∵ sin²θ + cos²θ = 1]