Question

hi dears

© if 2tanalpha = 3tanbita , then prove that tan( alpha - bita ) = sin2bita/ 5 - cos2bita

points: 60

Answer 1

2tanα=3tanβ
or, tanα=(3/2)tanβ
∴, tan(α-β)=(tanα-tanβ)/(1+tanα.tanβ)

={(3/2)tanβ-tanβ}/{1+(3/2)tanβ.tanβ}

={(3tanβ-2tanβ)/2}/{(2+3tan²β)/2}

=(tanβ/2)×2/(2+3tan²β)

=tanβ/(2+3tan²β)

=(sinβ/cosβ)/{2+3(sin²β/cos²β)}

=(sinβ/cosβ)/{(2cos²β+3sin²β)/cos²β}

=(sinβ/cosβ)×cos²β/{2cos²β+3(1-cos²β)}; [∵,sin²β+cos²β=1]

=sinβcosβ/(2cos²β+3-3cos²β)

=2sinβcosβ/2(3-cos²β)

=sin2β/(6-2cos²β)

=sin2β/(5+1-2cos²β)

=sin2β/{5-(2cos²β-1)}

=sin2β/(5-cos2β)

Hence Proved

Hope this helps you.