Question 21469
 tan({{{13pi/12}}})
 = tan({{{pi/12}}})

  Use {{{cos 2 x = 2 cos^2 x -1 = 1- 2 sin ^2x }}}
 = {{{sin (x/2)/ cos (x/2 ) }}}
 
We have {{{tan (x/2) = sqrt ((1-cos x)/(1+cos x) )}}}

Hence, {{{tan (pi/12) = sqrt((1-cos (pi/6) )/(1+cos (pi/6)))}}}

  = {{{ sqrt((1- sqrt(3)/2)/(1 + sqrt(3)/2))}}}
  = {{{ sqrt((1- sqrt(3)/2)^2/ (1/2)^2) }}}
 = {{{2 - sqrt(3)}}}


 Kenny