Question

tan（α-β）=sin2β 求证 2tan2β=tanα+tanβ

Answer 1

∵tan（α-β）=sin2β
tanα+tanβ
=tan[(α-β)+β]+tanβ
=[tan(α-β)+tanβ]/[1-tan(α-β)tanβ]+tanβ
=[sin2β+sinβ/cosβ)/[1-sin2β*sinβ/cosβ]+sinβ/cosβ
=[2sinβcos²β+sinβ]/(cosβ-2sin²βcosβ)+sinβ/cosβ
=sinβ(2cos²β+1)/[cosβ(1-2sin²β)+sinβ/cosβ
=sinβ/cosβ*[2cos²β+(1+cos2β)]/cos2β
=sinβ/cosβ*4cos²β/cos2β
=4sinβcosβ/cos2β
=2sin2β/cos2β
=2tan2β
即原等式成立