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2(tan a+tan b)
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已知tan(A-B/2)=2,tan(B-A/2)=-3 求:(1) tan[(A+B)/2] (2) tan(A+B)
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证明:(1+tan a+cot a)/(1+tan^2 a+tan a)-cot a/(1+ tan^2 a)=sin
已知三角形ABC中.2B=A+C,求tan(A/2)+tan(C/2)+根号3tan(A/2)*tan(C/2)的值
在三角形ABC中,已知A+C=2B,求Tan(A/2)+Tan(C/2)+√3Tan(A/2)×Tan(C/2)