已知sin(α+β)=5sinβ,则tan（α+β)*cotα=

sin(α+β)=5sinβ
sinacosb+cosasinb=5sinb
sinacosb=sinb(5-cosa)
cosb=sinb(5-cosa)/sina
cos^2b+sin^2b=1=sin^2b[(5-cosa)^2/sin^2a+1]=sin^2b[(26-10cosa)/sin^2a]
sinb=+-sina/√(26-10cosa),cosb=+-(5-cosa)/√(26-10cosa)
tan（α+β)*cotα=sin(a+b)cosa/[cos(a+b)sina]=5sinbcosa/[cosacosb-sinasinb)sina
=[5sinacosa/√(26-10cosa)]/{[cosa(5-cosa)/√(26-10cosa)-sin^2a/√(26-10cosa)]sina}
=[5cosa]/[cosa(5-cosa)-sin^2a]
=5cosa/(5cosa-cos^2a-sin^2a)
=1+1/(5cosa-1)
再问： sin(2α+β)=5sinβ,打错了
再答： sin(2α+β)=5sinβ sin(2a)cosb+cos(2a)sinb=5sinb sin(2a)cosb=sinb[5-cos(2a)] cosb=sinb[5-cos(2a)]/sin(2a) cos^2b+sin^2b=1=sin^2b{[5-cos(2a)]^2/sin^2(2a)+1}=sin^2b[26-10cos(2a)]/sin^2(2a) sinb=sin(2a)/√[26-10cos(2a)],cosb=[5-cos(2a)]/√[26-10cos(2a)] tan（α+β)*cotα=sin(a+b)cosa/[cos(a+b)sina] =(sinacosb+cosasinb)cosa/[cosacosb-sinasinb)sina] ={[sina[5-cos(2a)]/√[26-10cos(2a)]+cosasin(2a)/√[26-10cos(2a)]}cosa / {cosa[5-cos(2a)]/√[26-10cos(2a)]-sinasin(2a)/√[26-10cos(2a)]}sina ={[sina[5-cos(2a)]+cosasin(2a)}cosa / {cosa[5-cos(2a)]-sinasin(2a)}sina =[5-cos(2a)+2cos^2a]cosa / {cosa[5-cos(2a)]-sinasin(2a)} =[5-cos(2a)+2cos^2a] /[5-cos(2a)-2sin^2a] =[5-(2cos^a-1)+2cos^a]/[5-(1-2sin^2a)-2sin^2a] =(5+1)/(5-1) =3/2