- 三角求值问题在三角形ABC中,BC=a,CA=b,AB=C.若9
- 在三角形AB中,BC=a,CA=b,AB=C.若9a^2+9b^2-19c^2=0,求cotC/(cotA+cotB).
- 解:
cotC/(cotA+cotB)
=(cosC/sinC)/[(cosA/sinA)+(cosB/sinB)]
=(sinAsinBcosC)/(sinC)^2
=(ab/c^2)*[(a^2+b^2-c^2)/2ab]
=(a^2+b^2-c^2)/(2c^2)
=(9a^2+9b^2-9c^2)/(2×9c^2)
=(19c^2-9c^2)/(18c^2)
=5/9.