已知cos(α-π/6)=12/13,且π/6<α<π/2,求tanα

已知cos(α-π/6)=12/13,且π/6<α<π/2,求tanα

π/6<α<π/2
所以
0<α-π/6<π/3
而
cos(α-π/6)=12/13
所以
sin(α-π/6)=5/13
tan(α-π/6)=5/12
tanα=tan[(α-π/6)+π/6]
=[tan(α-π/6)+tanπ/6]/[1-tan(α-π/6)tanπ/6]
=[5/12+√3/3]/[1-5/12×√3/3]
=（15+12√3）/(36-5√3)

α和β太难打，所以我用a和b来代替，楼主将就下

tan(a+b) =2tan[(a+b)/2]/{1-{tan[(a+b)/2]}^2} =√6/(1-6/4) =-2√6 tana+tanb =tan(a+b)*(1-tanatanb) =-2√6*(1-13/7) =(12√6)/7 (tana-tanb)^2 =(tana+tanb)^2-4tanatanb =864/49-52/7 =500/49 tan(a-b)=(tana-tanb)/(1+tanatanb) [tan(a-b)]^2 =(tana-tanb)^2/(1+tanatanb)^2 =(500/49)/(1+13/7)^2 =5/4 1+[tan(a-b)]^2=1/[cos(a-b)]^2 1+5/4=1/[cos(a-b)]^2 cos(a-b)=2/3 tana和tanb均为正数，tan[(a+b)/2]也为正数且大于1，所以a、b同象限，cos(a-b)为正

不懂可以追问，满意请采纳下

cos(α-π/6)=12/13 π/6<α<π/2 0< α-π/6<π/3 sin(α-π/6)=√(1-144/169)=5/13 sinα=sin[(α-π/6)+π/6] =sin(α-π/6)cosπ/6+cos(α-π/6)sinπ/6 =5/13*√3/2+12/13*1/2=(5√3+12)/26 cosα=cos[(α-π/6)+π/6] =12/13*√3/2-5/13*1/2=(12√3-5)/26 tanα=sinα/cosα =(5√3+12)/(12√3-5) =(240+169√3)/407

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