在△ABC中…已知sinA=cosB*cosC求证tgB+tgC=1

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在△ABC中…已知sinA=cosB*cosC求证tgB+tgC=1
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在△ABC中…已知sinA=cosB*cosC求证tgB+tgC=1
在△ABC中…已知sinA=cosB*cosC
求证tgB+tgC=1

在△ABC中…已知sinA=cosB*cosC求证tgB+tgC=1
tgB+tgC = sinB/cosB + sinC/cosC
=(sinB·cosC + cosB·sinC)/(cosB·cosC)
=sin(B+C)/(cosB·cosC)
=sin(π-A)/(cosB·cosC)
=sinA/(cosB·cosC)
=sinA/ sinA
= 1

cosBcosC=sinA=sin〔派-(B+C)〕=sin(B+C)=sinBcosC+cosBsinC 即是sinBcosC+cosBsinC=cosBcosC 等式两边同时除以cosBcosC得〔sinBcosC+cosBsinC〕/cosBcosC=1 等式左边整理即得所证结果(手机回的,有些东西写不出来,见谅)