已知 tan(a-c)/tana +sin^b/sin^a=1,求证tan^b=tana*tanc
热心网友
tan(a-c)/tana +sin^b/sin^a=1== sin^b/sin^a = 1 - tan(a-c)/tana = 1 - (tga-tgc)/[tha(1+tgatgc)]= tgc[1+(tga)^2]/[tga(1+tgatgc)]又: sin^b/sin^a = (tgb)^2/[1+(tgb)^2] * [1+(tga)^2]/(tga)^2因此: (tgb)^2/[1+(tgb)^2]*[1+(tga)^2]/(tga)^2 = tgc[1+(tga)^2]/[tga(1+tgatgc)](tgb)^2/[1+(tgb)^2] = tgatgc/(1+tgatgc)因此: (tgb)^2 = tgatgc