If a sin θ + b cos θ = c, what is/are the values of (a cos θ – b

1.	If a sin θ + b cos θ = c, what is/are the values of (a cos θ – b sin θ)?(a) c – a + b (b) c – b + a (c) \(±\sqrt{a^2+b^2-c^2}\)(d) \(±\sqrt{c^2+b^2-a^2}\)
Answer» (c) \(±\sqrt{c^2+b^2-a^2}\) Given, a sin θ – b cos θ = c (a cos θ – b sin θ)² = a² cos²θ – 2ab cos θ sin θ + b² sin²θ = a² (1 – sin²θ) – 2(a sin θ) (b cos θ) + b² (1 – cos²θ) = a² – a² sin²θ – 2 (a sin θ) (b cos θ) + b² – b² cos²θ = a² + b² – [a² sin²θ + 2 (a sin θ) (b cos θ) + b² cos²θ] = a² + b² – (a sin θ + b cos θ)² = a² + b² – c² ⇒ (a cos θ – b sin θ) = \(±\sqrt{c^2+b^2-a^2}\)

If a sin θ + b cos θ = c, what is/are the values of (a cos θ – b sin θ)?(a) c – a + b (b) c – b + a (c) \(±\sqrt{a^2+b^2-c^2}\)(d) \(±\sqrt{c^2+b^2-a^2}\)

Answer»

(c) \(±\sqrt{c^2+b^2-a^2}\)

Given, a sin θ – b cos θ = c

(a cos θ – b sin θ)²

= a² cos²θ – 2ab cos θ sin θ + b² sin²θ

= a² (1 – sin²θ) – 2(a sin θ) (b cos θ) + b² (1 – cos²θ)

= a² – a² sin²θ – 2 (a sin θ) (b cos θ) + b² – b² cos²θ

= a² + b² – [a² sin²θ + 2 (a sin θ) (b cos θ) + b² cos²θ]

= a² + b² – (a sin θ + b cos θ)² = a² + b² – c²

⇒ (a cos θ – b sin θ) = \(±\sqrt{c^2+b^2-a^2}\)