tan(arctan1/5+arctan3)=
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tan(arctan1/5+arctan3)=tan(arctan1/5+arctan3)=tan(arctan1/5+arctan3)=tan(A+B)=(tanA+tanB)/(1-tanAtan
tan(arctan1/5+arctan3)=
tan(arctan1/5+arctan3)=
tan(arctan1/5+arctan3)=
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
tan(arctan1/5+arctan3)=(1/5+3)/(1-3/5)=8
设a=arctan1/5,b=arctan3
则tana=1/5,tanb=3
tan(arctan1/5+arctan3)
=tan(a+b)
=(tana+tanb)/(1-tana·tanb)
=(1/5 + 3)/(1- 1/5 ×3)
=8
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反三角函数2个问题,急求1:tan(arctan1/5+arctan3)2:cos【2arctan(3/4)】都是求值
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