Question

if a cos cube theta + 3 Alpha sin square theta cos theta = 2 m and a sin cube theta + 3 is sin theta cos square theta = 2 and prove that n + and 2 by 3

Answer 1

Given: 2 sin θ + 3 cos θ = 2
Consider (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2 = 4sin2 θ + 9 cos2 θ + 12sin θ cos θ + 9 sin2 θ + 4 cos2 θ – 12 sin θ cos θ

⇒ (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2 = 13sin2 θ + 13 cos2 θ

⇒ (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2 = 13(sin2 θ + cos2 θ)

⇒ (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2 = 13

⇒ (2)2 + (3 sin θ – 2 cos θ)2 = 13

⇒ (3 sin θ – 2 cos θ)2 = 13 – 4

⇒ (3 sin θ – 2 cos θ)2 = 9

⇒ (3 sin θ – 2 cos θ) = ± 3

Hence, proved.