\[\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \]
Test value- For a=3 b=4 c=5
we get s=(a+b+c)/2 = 6
and Tan(A/2)= 0.3333
where A is the angle of triangle opposite to side AB of length a
and s is semiperimeter which is equal to (a+b+c)/2
a , b ,c are the lenghts of triangle ABC
Note that a is the side opposite to the angle A
b is the side opposite to the angle B of the triangle
c is the side opposite to the angle C of the triangle
Boundary condition- a . b ,c must be positive and sum of any two sides of triangle must be greater than the third one
