in a triangle ABC where a,b,c are the sides opposite to the angles A,B,C respectively
AC is double of AB, then find the value of $\displaystyle {cot\:\dfrac{A}{2} }{cot\:\dfrac{B-C}{2} }$
$\displaystyle {\dfrac{AB}{AC} = \dfrac{1}{2} = \dfrac{\sin{C}}{\sin{B}}$
= $\displaystyle {\dfrac{(2+1)}{(2-1)} = \dfrac{(sinB +sinC)}{(sinB - sinC)}}$
= $\displaystyle {\dfrac{2sin\dfrac{B+C}{2}cos\dfrac{B-C}{2}}{2cos\dfrac{B+C}{2}sin\dfrac{B-C}{2}}}$
Since (B+C)/2 = π/2 - A/2, sin[(B+C)/2] = cos(A/2) and cos[(B+C)/2] = sin(A/2).
Substtitue these values in the above equation and simplify.