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

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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-...