Question

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 1

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]