- 三角问题三角形ABC中,a^2+b^2=6c^2,则(cotA+
- 三角形AB中,a^2+b^2=6c^2,则(cotA+cotB)×tanC的值等于______
- (cotA+cotB)tanC
=(cosA/sinA+cosB/sinB)tanC 同角三角函数的关系
=(cosAsinB+sinAcosB)/(sinAsinB)*sinC/cosC
=sin(A+B)/(sinAsinB)*sinC/cosC) 二角和的正弦
=sinCsinC/(sinAsinB)*1/cosC
=2RsinC*2RsinC/(2RsinA*2RsinB)*1/[(a^2+b^2-c^2)/(2ab)]
=c^2/(ab)*2ab/(a^2+b^2-c^2) 正弦定理、余弦定理
=2c^2/(a^2+b^2-c^2)
=2c^2/(6c^2-c^2) 已知条件
=2c^2/(5c^2)
=2/5.