已知△ABC的三内角A,B,C所对三边分别为a,b,c,且sin(π/4+A)=7√2/10,0＜A＜π/4（1）求ta
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sin(π/4+A)=√2/2(sinA+cosA)=7√2/10;

(sinA+cosA)=7/5;

(sinA+cosA)=[2sin(A/2)cos(A/2)+cos(A/2)^2-sin(A/2)^2]/[sin(A/2)^2+cos(A/2)^2];

上下除以cos(A/2)^2得：

(sinA+cosA)=[2tan(A/2)+1-tan(A/2)^2]/[tan(A/2)^2+1]=7/5;

7tan(A/2)^2+7=10tan(A/2)+5-5tan(A/2)^2;

12tan(A/2)^2-10tan(A/2)+2=0;

6tan(A/2)^2-5tan(A/2)+1=0;

tan(A/2)=1/3或1/2;

tanA=[2tan(A/2)]/[1-tan(A/2)^2]=3/4或4/3;

而0＜A＜π/4；所以tanA

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