求证：在三角形ABC中,三角形的面积等于(a*2)/[2(cotB+cotC)]

因该是a^2 吧
a^2 / 2(cosB/sinB + cosC/sinC)
=a^2 / 2(cosB*sinC + cosC*sinB/sinB*sinC)
上下乘sinBsinC,得
a^2*sinB*sinC / 2(cosB*sinC + sinB*cosC)
=a^2*sinB*sinC / 2sin(B+C)
=a^2*sinB*sinC / 2sinA
=a^2*sinB*sinC / 2(sinB/b)
=a*b*sinC / 2
=S△ABC
=左边
∴原命题成立