问题 填空题
0<y≤x<
π
2
,且tanx=3tany,则x-y的最大值为______.
答案

因为0<y≤x<

π
2
,x-y∈(0,
π
2
),且tanx=3tany,

所以tan(x-y)=

tanx-tany
1+tanxtany

=

2tany
1+3tan2y

=

2
1
 tany
+3tan y

2
2
1
tany
•3tany

=

3
3

=tan

π
6
,当且仅当3tan2y=1时取等号,

∴x-y的最大值为:

π
6

故答案为:

π
6

解答题
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