作业帮tan(π/8)+1/(tan(π/12))值,要过程
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tan(π/8)+1/(tan(π/12))值,要过程
tan(π/8)+1/(tan(π/12))值,要过程
tan(π/8)+1/(tan(π/12))值,要过程
tan(π/4)=2tan(π/8)/(1-tan(π/8)^2)=1
tan(π/6)=2tan(π/12)/(1-tan(π/12)^2)=(根号3)/3
解二次方程
因为tan(π/8),tan(π/12)＞0
求根公式取正
tan(π/8)=（根号2）-1
tan(π/12)=2-根号3
所以原式=（根号2)-1+1/(2-根号3)
=(根号2)-1+2+根号3
=根号3+根号2+1
tan(π/12)-(1/tan(π/12))
化简：tan(π/8)+1/tan(π/12)
tan(π/8)+1/(tan(π/12))值,要过程
(1-tan^2π/12)/(tanπ/12)化简
计算 tanπ/12+1/tanπ/12
计算tanπ/12 / (1-tan^2π/12)=
tan(π/12)+1/tan(π/12)=啥?
tanπ/12-1/tanπ/12=
(tan(5π/8)*tan(π/8))为什么等于-1
tanπ/8/(1-tan^2π/8)
tanπ/8-1/tanπ/8是（tanπ/8）-〔1/（tanπ/8）〕。
三角函数算值题1-tan^2(π/8) /tan(π/8)=tan(15)/1-tan^2(15)=
化简：[tan(5/4)π+tan(5/12)π]/[1-tan(5/12)π]
化简：[tan（5π）/4+tan(5π)/12]/[1-tan(5π)/12]
求tan(π/8)
tan(π/8)=?
（tanπ/4+tanα)/(1-tanπ/4tanα)怎么解
利用正切函数单调性比较 函数值大小tan [(75/11)π]&tan [(-58/11)π]这样可不可以 下面算式有错吗tan[(75/11)π] = tan[(9/11)π] =tan[(1-2/11)π] = -tan 2/11 πtan [(-58/11)π] = tan [-(8/11)π] = -tan[(1-3/11)π] = tan 3/11 π