JEE Main 2019MathematicsTrigonometric Ratios & IdentitiesMediumMCQ

JEE Main 2019 — Trigonometric Ratios & Identities Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

The value of cos210°–cos10° cos50°+cos250° is

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Show full solutionCorrect option: A
Correct answer
A34

Step-by-step explanation

cos210°+ cos250°-cos10°cos50°

Using, cos2x=2cos2x-1, ⇒cos2x=1+cos2x2, we get

=121+cos20°+1+cos100°-2cos50°cos10°

=122+cos100°+cos20°-cos60°-cos40°

=122+cos100°+cos20°-12-cos40°

Using, cosC+cosD=2cosC+D2cosC-D2, we get

=1232+2cos⁡60°cos⁡40°-cos⁡40°

=1232+2×12×cos40°-cos40°

=1232+cos40°-cos40°

=12×32=34.

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About this question

This is a previous-year question from JEE Main 2019, covering the Trigonometric Ratios & Identities 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.