求解一道三角函数体已知sin(2x+y)=5siny,求证2tan(x+y)=3tanx.

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求解一道三角函数体
已知sin(2x+y)=5siny,求证2tan(x+y)=3tanx.

别人提问是需要有详细解答过程的：
sin[(x+y)+x]=5sin[(x+y)-x]
sin(x+y)·cosx＋cos(x+y)·sinx＝5·sin(x+y)·cosx－5·cos(x+y)·sinx
4·sin(x+y)·cosx＝6·cos(x+y)·sinx
两边同除以cos(x+y)·cosx得2·tan(x+y)=3·tanx

已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx 已知：3sinY=sin(2X+Y),求证tan(X+Y)=2tanX 3sinX=sin(2X+Y) 求证：tan(X+Y)=2tanX tan(x+y)=2tanx(x,x+y≠kπ+π/2,k∈Z),证3siny=sin(2x+y) 已知cos(x+y)cosy+sin(x+y)siny=4/5,求tanx的值已知sinx+siny=2sin(x+y)≠0,则tanx／2tany／2等于是三角函数和差化积部分的已知sin（x+y）=1,求证：tan（2x+3y）=tany 求证sinx+siny=2sin(x+y)/2*cos(x-y)/2 求证|sinx-siny|=|2sin[(x-y)/2]cos[(x+y)/2]| 求证：sin(2x+y)/sinx-2cos(x+y)=siny/sinx 求函数y=（tan²x-tanx+1）除以（tan²x+tanx+1）的值域已知函数y=sin（2 已知sin(x+y)=1,求证：tan(2x+y)+tany=0