If 2y . cos θ - x . sin θ = 0 and 2x . sec θ - y . cosec θ = 3, what is the value of x² + 4y²?
Given, 2y . cos θ - x . sin θ = 0 ⇒ 2y . cos θ = x . sin θ … (i) Also given, 2x . sec θ - y . cosec θ = 3 ⇒ 2x/cos θ - y/sin θ = 3 ⇒ 2x . sin θ - y . cos θ = 3 . sin θ . cos θ ⇒ 2 * 2y . cos θ - y . cos θ = 3 . sin θ . cos θ [from (i)] ⇒ 3y . cos θ = 3 . sin θ . cos θ ⇒ y = sin θ ⇒ x = 2 . cos θ [substituting the value of y in (i)] ∴ x² + 4y² = (2 . cos θ )² + 4 (sin θ )² = 4 cos² θ + 4 sin² θ = 4 * (cos² θ + sin² θ ) = 4 * 1 = 4
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