已知tan[(α+β)/2]=(根号6）/2,tanαtanβ=13/7 求cos(α-β的值

tan(a+b) =2tan[(a+b)/2]/{1-{tan[(a+b)/2]}^2} =√6/(1-6/4) =-2√6 tana+tanb =tan(a+b)*(1-tanatanb) =-2√6*(1-13/7) =(12√6)/7 (tana-tanb)^2 =(tana+tanb)^2-4tanatanb =864/49-52/7 =500/49 tan(a-b)=(tana-tanb)/(1+tanatanb) [tan(a-b)]^2 =(tana-tanb)^2/(1+tanatanb)^2 =(500/49)/(1+13/7)^2 =5/4 1+[tan(a-b)]^2=1/[cos(a-b)]^2 1+5/4=1/[cos(a-b)]^2 cos(a-b)=2/3 tana和tanb均为正数,tan[(a+b)/2]也为正数且大于1,所以a、b同象限,cos(a-b)为正