求tan[(x1+x2)/2]及tanx1 * tanx2设x1,x2为以x为变量方程5cosx+6sinx=7的两根,不解x1,x2,求1)tan[(x1+x2)/2]及tanx1 * tanx2

tan(A/2+B/2)=(sinA+sinB)/(cosA+cosB)---------
5cosx+6sinx=7,
7-5cosx=6sinx,
49-70cosx+25cosx cosx=36sinx sinx,
13-70cosx=-36+36sinx sinx-25cosx cosx,
13-70cosx=-61cosx cosx,
13-70cosx+61cosx cosx=0,
cosx1+cosx2=70/61,
cosx1 * cosx2=13/61,
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5cosx+6sinx=7,
7-6sinx=5cosx
49-84sinx+36sinx sinx=25cosx cosx
24-84sinx+61sinx sinx=0
sinx1+sinx2=84/61
sinx1 * sinx2 =24/61
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tanx1 * tanx2=(sinx1 * sinx2)/(cosx1 * cosx2)=24/13,
tan[(x1+x2)/2]=(sinx1+sinx2)/(cosx1+cosx2)=84/70=6/5;