If (4sinθ + 3cosθ)2 – (2sinθ + cosθ)2 = 20, then, what is the value of tanθ ?

If (4sinθ + 3cosθ)2 – (2sinθ + cosθ)2 = 20, then, what is the value of tanθ ? Correct Answer 3/2 or 1

Given:

(4sin θ + 3cos θ)² – (2sin θ + cos θ)² = 20

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

(4sin θ + 3cos θ)² – (2sin θ + cos θ)² = 20

⇒ (16sin²θ + 9cos²θ + 24sin θ cos θ) – (4sin²θ + cos²θ + 4sin θ cos θ) = 20

⇒ 12sin²θ + 20sin θ cos θ + 8cos²θ = 20

⇒ 3sin²θ + 5sin θ cos θ + 2cos²θ = 5

⇒ 3sin²θ + 5sin θ cos θ + 2cos²θ = 5 × (sin²θ + cos²θ)

⇒ 2sin²θ – 5sin θ cos θ + 3cos²θ = 0

⇒ (2sin²θ – 2sin θ cos θ – 3sin θ cos θ + 3cos²θ) = 0

⇒ 2sin θ(sin θ – cos θ) – 3cos θ(sin θ – cos θ) = 0

⇒ (sin θ – cos θ) (2sin θ – 3cos θ) = 0

⇒ (sin θ – cos θ) = 0 or (2sin θ – 3cos θ) = 0

⇒ tanθ = 1 or tanθ = 3/2