If 2ycos θ = x sin θ and 2xsec θ – y cosec θ = 3, then the value of x2 + 4y2 is

If 2ycos θ = x sin θ and 2xsec θ – y cosec θ = 3, then the value of x² + 4y² is

⇒ 2ycos θ = x sin θ      ----(1)

⇒ Now squaring on both side

⇒ 4y²cos² θ = x² sin² θ      ----(2)

⇒ 2xsec θ – y cosec θ = 3

⇒ 2x/cos θ – y/sin θ = 3

⇒ 2x sin θ – y cos θ = 3sin θcos θ

⇒ from equation 1 put ycos θ = xsin θ/2

⇒ 2x sin θ – x sin θ/2 = 3 sin θ cos θ

⇒ 3 sin θ = 6 sin θ cos θ

⇒ x = 2cos θ      ----(3)

⇒ Now, x² + 4y²

⇒ (2cos θ)² + x² sin² θ/cos² θ

⇒ 4cos² θ + 4 cos² θ sin² θ/cos² θ

∴ 4(cos² θ + sin² θ) = 4

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