JEE Main 2024MathematicsTrigonometric Ratios & IdentitiesMediumMCQ

JEE Main 2024 — Trigonometric Ratios & Identities Question with Solution

JEE Main 2024 (01 Feb Shift 1)

Question

If tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112,0<A,B,C<π2, then A+B is equal to:

Choose an option

Show full solutionCorrect option: A
Correct answer
AC

Step-by-step explanation

Given, tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112=1+x+x2xx

We know that,

tanA+B=tanA+tanB1−tanAtanB

⇒tanA+B=1xx2+x+1+xx2+x+11−1x2+x+1

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=x2+x+1xx

⇒tanA+B=tanC

⇒A+B=C

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About this question

This is a previous-year question from JEE Main 2024, 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.