题目是：已知a>1方程x^2+4ax+3a+1=0的两根为tanc,tand 则tan[(c+d)/2]?
tan(c+d)=(tanc+tand)/(1-tanctand)=-4a/(1-3a-1)=4/3
tan2a=2tana/(1-tana^2)
设tan[(c+d)/2=x
2x/(1-x^2)=4/3
=>4(1-x^2)=6x
=>x=1/2,-2
=>tan[(c+d)/2=1/2 或者-2