△ABC中,A＝π/3,并且tan(π/4＋B）等于﹣3,求tanC
问题描述：
△ABC中,A＝π/3,并且tan(π/4＋B）等于﹣3,求tanC
答
tan(π/4＋B)=﹣3 (1+tanB)/(1-tanB)=-3 tanB=2
tanC=[π-（A+B)]=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)
=-(tanπ/3+tanB)/(1-tanπ/3anB)
=-(√3+2)/(1- 2√3)
=(8+5√3)/11
答
tan(π/4＋B)=[tan(π/4)＋tanB]/[1－tan(π/4)tanB]=－3
得：
tanB=2
因为：A=π/3、tanB=2
则：
tanC
=tan[π－(A＋B)]
=－tan(A＋B)
=－tan(π/3＋B)
=－[tan(π/3)＋tanB]/[1－tan(π/3)tanB]
=－[√3＋2]/[1－2√3]
=(8＋5√3)/11