【已知:5cos(a-b/2)+7cos(b/2)=0求:tan(a/2)*tan((a-b)/2)已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].由5cos(a-b/2)+7cos(b/2)=0得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=05{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=010[cos^2(a/2)cos(b】

　　已知:5cos(a-b/2)+7cos(b/2)=0

　　求:tan(a/2)*tan((a-b)/2)

　　已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].

　　由5cos(a-b/2)+7cos(b/2)=0

　　得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=0

　　5{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=0

　　10[cos^2(a/2)cos(b/2)+10sin(a/2)cos(a/2)sin(b/2)+2cos(b/2)=0

　　7cos(b/2)——>2cos(b/2)这为什么会从7跑到2的?

　　用2cos(b/2)除两边得:

　　5cos^2(a/2)+5sin(a/2)cos(a/2)tan(b/2)+1=0

　　∴tan(b/2)=-[1+5cos^2(a/2)]/5sin(a/2)cos(a/2)

　　=-[sin^2(a/2)+6cos^2(a/2)]/5sin(a/2)cos(a/2)

　　=-[tan^2(a/2)+6]/5tan(a/2).

　　于是tan(a/2)tan[(a-b)/2]=tan(a/2)[tan(a/2)-tan(b/2)]/[1+tan(a/2)tan(b/2],将tan(b/2)的表达式代入,化简得:

　　tan(a/2)tan[(a-b)/2]=-6.