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JEE Main201910 Apr 2019Evening ShiftMathematicsInverse Trigonometric FunctionsActual

If cos - 1 x - cos – 1 y 2 = α , where - 1 ≤ x ≤ 1 , - 2 ≤ y ≤ 2 , x ≤ y 2 , then for all x , y , 4 x 2 - 4 x y cos α + y 2 is equal to :

Options

  1. A4 cos 2 α + 2 x 2 y 2
  2. B4 sin 2 α - 2 x 2 y 2
  3. C2 sin 2 α
  4. D4 sin 2 α

Correct answer

D. 4 sin 2 α

Step-by-step solution

Given cos - 1 x - cos - 1 y 2 = α cos - 1 A - cos - 1 B = cos - 1 A B + 1 - A 2 1 - B 2 , ⇒ cos - 1 x y 2 + 1 - x 2 × 1 - y 2 4 = α ⇒ x y 2 + 1 - x 2 × 1 - y 2 4 = cos α ⇒ cos α - x y 2 = 1 - x 2 × 1 - y 2 4 Squaring both sides, we get cos 2 α + x 2 y 2 4 - x y cos α = 1 - x 2 1 - y 2 4 ⇒ cos 2 α + x 2 y 2 4 - x y cos α = 1 - x 2 - y 2 4 + x 2 y 2 4 ⇒ x 2 - x y cos α + y 2 4 = 1 - cos 2 α ⇒ x 2 - x y cos α + y

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