If 2tanα=3tanβ than prove that tanβ/(2+3tan^2β)

Steps to prove tan(α-β)=sin2β/(5-cos2β)

L.H.S

=tan(α-β)

=(tanα-tanβ)/(1+tanα.tanβ)

={(3/2)tanβ-tanβ}/{1+(3/2)tanβ.tanβ}

={(3tanβ-2tanβ)/2}/(2+3tan^2β)

=tanβ/(2+3tan^2β)

R.H.S

=sin2β/(5-cos2β)

={2tanβ/(1+tan^2β)}/{5-(1-tan^2β)/(1+tan^2β)}

={2tanβ/(1+tan^2β)}/[{5(1+tan^2β)-(1-tan^2β)}/(1+tan^2β)

=2tanβ/(4+6tan^2β)

=tanβ/(2+3tan^2β)

Hence L.H.S=R.H.S

tanβ/(2+3tan^2β) is proved.

Detail Information:-

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