tan(-23π/6)=

tan(-23π/6)= 求tan(-6分之23π) tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β) 求值：tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)= tan(-6分之31π)=什么? tan(-23派/6) tan（π/6-α）+tan（π/6+α）+√3tan（π/6-α）tan（π/6+α）=? 化简tan2αtan(π/6-α)+tan2αtan(π/3-α)+tan(π/6-α)tan(π/3-α)=?请写下过程, tan=3,π tan(π/8)=? tan?=1/2 tan?=1/3 tan?=1/4 tan?=1/5 tan?=1/6 tan(π/6-2)=根号3/2,求tan(5π/6+2)的值 tan(π/7+α)=5,则tan(6π/7-α)= 已知tan[(π/7)+a]=5,则tan[(6π/7)-a]= 若3^x=tanπ/6,则tan（π+2x）=? tanα=2则tan（α-π/6）=5根号3-8, 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? 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