JEE Main 2019MathematicsInverse Trigonometric FunctionsMediumMCQ

JEE Main 2019 — Inverse Trigonometric Functions Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

If cos-1x-cos–1y2=α, where -1≤x≤1, -2≤y≤2, x≤y2, then for all x, y, 4x2-4xycosα+y2 is equal to :

Choose an option

Show full solutionCorrect option: D
Correct answer
D4sin2α

Step-by-step explanation

Given cos-1x-cos-1y2=α

cos-1A-cos-1B=cos-1AB+1-A21-B2,

⇒cos-1xy2+1-x2×1-y24=α

⇒xy2+1-x2×1-y24=cosα

⇒cosα-xy2=1-x2×1-y24

Squaring both sides, we get
cos2α+x2y24-xycosα=1-x21-y24

⇒cos2α+x2y24-xycosα=1-x2-y24+x2y24

⇒x2-xycosα+y24=1-cos2α

⇒x2-xycosα+y24=sin2α

⇒4x2-4xycosα+y2=4sin2α.

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About this question

This is a previous-year question from JEE Main 2019, covering the Inverse Trigonometric Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.