JEE Main 2023MathematicsBasic of MathematicsMediumMCQ

JEE Main 2023 — Basic of Mathematics Question with Solution

JEE Main 2023 (30 Jan Shift 1)

Question

If the solution of the equation logcosxcotx+4logsinxtanx=1, x∈0,π2 is sin-1α+β2, where α,β are integers, then α+β is equal to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D4

Step-by-step explanation

Given:

logcosxcotx+4logsinxtanx=1, x∈0,π2

⇒lncosx-lnsinxlncosx+4lnsinx-lncosxlnsinx=1

⇒1-lnsinxlncosx+41-lncosxlnsinx=1

⇒lnsinxlncosx+4lncosxlnsinx-4=0

⇒(lnsinx)2-4(lnsinx)(lncosx)+4(lncosx)2=0

⇒lnsinx-2lncosx2=0

⇒lnsinx=2lncosx

⇒lnsinx=lncos2x

⇒cos2x=sinx

⇒1-sin2x=sinx

⇒sin2x+sinx-1=0

⇒sinx=-1+52

So, α=-1, β=5

∴α+β=4

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About this question

This is a previous-year question from JEE Main 2023, covering the Basic of Mathematics 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.