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JEE Main 2019 · 10 April · Shift II

JEE Main 2019 Mathematics question on Inverse Trigonometric Functions

If cos⁡−1x−cos⁡−1y2=α\cos^{-1}x - \cos^{-1}\dfrac{y}{2} = \alpha, where −1≤x≤1-1 \leq x \leq 1, −2≤y≤2-2 \leq y \leq 2, x≤y2x \leq \dfrac{y}{2}, then for all xx, yy, 4x2−4xycos⁡α+y24x^2 - 4xy\cos\alpha + y^2 is equal to
  1. A.
    2sin⁡2α2\sin^2\alpha
  2. B.
    4cos⁡2α+2x2y24\cos^2\alpha + 2x^2y^2
  3. C.
    4sin⁡2α4\sin^2\alpha
  4. D.
    4sin⁡2α−2x2y24\sin^2\alpha - 2x^2y^2
Show answer and solution

Correct option: C

Solution

Let A = cos⁻¹x and B = cos⁻¹(y/2), so A - B = α.

Then cos(A - B) = cosA·cosB + sinA·sinB gives:
x·(y/2) + √(1-x²)·√(1-y²/4) = cosα

So √(1-x²)·√(1-y²/4) = cosα - xy/2.

Squaring: (1-x²)(1-y²/4) = cos²α - xy·cosα + x²y²/4

Expanding left side: 1 - x² - y²/4 + x²y²/4 = cos²α - xy·cosα + x²y²/4

Simplifying: 1 - x² - y²/4 = cos²α - xy·cosα

Rearranging: 4 - 4x² - y² = 4cos²α - 4xy·cosα

So: 4x² - 4xy·cosα + y² = 4 - 4cos²α = 4sin²α.

[JEE Main 2019 · 10 April · Shift II]

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