Rishi Sharma
Last Activity: 4 Years ago
Dear Student,
Please find below the solution to your problem.
If 2tanA=3tanB or tanA=(3/2).tanB
Prove that tan(A-B)=sin2B/(5-cos2B)
LHS.
tan(A-B)=(tanA-tanB)/(1+tanA.tanB), put tanA=(3/2).tanB.
=[(3/2)tanB-tanB]/[1+(3/2)tan^2B]
=[(1/2).tanB]/[(2cos^2B+3sin^2B)/2cos^2B].
=[(sinB/2cosB)×2cos^2B]/[2cos^2B+3sin^2B].
=[sinBcosB]/(2cos^2B+3sin^2B)
=(sin2B)/2.(2cos^2B+3sin^2B
=(sin2B)/(4cos^2B+6sin^2B)
=(sin2B)/[4(1-sin^2B)+6sin^2B]
=(sin2B)/[4–4sin^2B+6sin^2B]
=(sin2B)/[4+2sin^2B) , put 2sin^2B=1-cos2B
=(sin2B)/[4+1-cos2B]
=(sin2B)/(5-cos2B) , Proved.
Thanks and Regards