tan12分之π-(tan12分之π)分之1,tan12分之π-（tan12分之π）分之1,

Question

tan12分之π-(tan12分之π)分之1,
tan12分之π-（tan12分之π）分之1,

Answer 1

tan12分之π-（tan12分之π）分之1

= sin(π/12)/cos(π/12)-cos(π/12)/sin(π/12)
=[sin^2(π/12)-cos^2(π/12)]/[cos(π/12)sin(π/12)]
=-cos(2π/12)/[1/2sin(2π/12)]
=(-√3/2)/(1/4)
=-2√3