化简 tan((π/12)+1)/((π/12)-1) 求指导,
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/10 03:02:46
x){3<
Fچ0³MzڟߣcTO;2xh|~=/>EV6TUW@U ڢh8&ڴR(" #Du2/.H̳y
(1-tan^2π/12)/(tanπ/12)化简
化简:tan(π/8)+1/tan(π/12)
tan(π/12)-(1/tan(π/12))
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]
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(π/8)+1/(tan(π/12))值,要过程
化简(tan5π/4+tan5π/12)/(1-tan5π/12)(tan 5π/4 + tan 5π/12)/(1-tan 5π/12) =(tan π/4 + tan π/6+π/4 )/(1-tan π/4*tan π/6+π/4 ) =tan(π/4 + π/6+π/4 )=tan(π/2+π/6) 请问我这样算错在哪呢?tan π/2不是无意义吗?
(tanπ/4+tanα)/(1-tanπ/4tanα)怎么解
化简 tan((π/12)+1)/((π/12)-1) 求指导,
计算:1-tan(5π/12)tan(π/4) ------------------------------- tan(5π/12)+tan(π/4)那条横杠是分数线
化简(tan@+1比tan@)cos@的平方等于 (用特值法)例三分之π1.tan@ 2.sin@3cos@ 4.1比tan@
tan( x/2+π/4)+tan(x/2-π/4 )=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x
化简tan+1/tan
求证(1)1+tanθ/1-tanθ=tan(π/4+θ) (2)1-tanθ/1+tanθ=tan(π/4-θ)