Question

高中数学！！！

若tan2x=3tan（x-y）=3，则tan（x+y）=

Answer 1

tan2x=tan<x+y）+（x-y）> tan(x-y)=1

={tan(x+y) + tan(x-y) } / {1-tan(x+y)tan(x-y)}

=1+tan(x+y)/1-tan(x+y)=3
解得 tan(x+y)=1/2

Answer 2

tg(2x)= tg(x+y+ x-y)=[tg(x+y)+tg(x-y)]/[1-tg(x+y)*tg(x-y)]= [tg(x+y)+1]/[1-tg(x+y)*1]=3
解得tg(x+y)+1 = 3-3tg(x+y)
tg(x+y)=1/2

Answer 3

tan2x=2tanx/1-tanx^2

3tan(x-y)=3[tanx-tany/1+tanxtany]=3

tan(x-y)=[tanx-tany/1+tanxtany]=1

tan(x+y)=[tanx+tany/1-tanxtany]

所以tan(x+y)=-tan(x-y)=-1

Answer 4

答案是0.5

Answer 5

答案是1/2

Answer 6