已知asin(γ+α)=bsin(γ+β),求证tanγ=bsinβ-asinα/acosα-bcosβ
问题描述：
已知asin(γ+α)=bsin(γ+β),求证tanγ=bsinβ-asinα/acosα-bcosβ
答
a(sin ycos a+sin acos y)=b(sin ycos b+sin bcos y)asin ycos a-bsin ycos b=bsin bcos y-asin acos ysin y(acos a-bcos b)=cos y(bsin b-asin a)sin y/cos y=(bsin b-asin a)/(acos a-bcos b)=tan y得证