trigonometry---1

a1)
a/sinA = b/sinB = k
a = ksinA , b = ksinB
a - b = ksinA - ksinB
a - b = k(sinA - sinB)
a2)
cos(A+B)/2 cot(C/2)
= cos[(180-C)/2] / tan(C/2)
= cos(90 - C/2) / tan(C/2)
= sin(C/2) / tan(C/2)
= cos(C/2)
= sin(90 - C/2)
= sin[(180 - C)/2]
= sin [(A+B)/2]




b1) By a2) , cot(C/2) = sin[(A+B)/2] / cos[(A+B)/2] , so
(a-b) cot C/2
= [k(sinA - sinB)] * sin[(A+B)/2] / cos[(A+B)/2]
= k * 2cos[(A+B)/2] sin[(A-B)/2] * sin[(A+B)/2] / cos[(A+B)/2]
= k * 2sin[(A+B)/2] * sin[(A-B)/2]
= - k (- 2sin[(A+B)/2] * sin[(A-B)/2])
= - k (cosA - cosB)
b2) By b1) , we have
(b-c) cot A/2 = - k(cosB - cosC) ;
(c-a) cot B/2 = - k(cosC - cosA)
So (a-b) cot C/2 + (b-c) cot A/2 + (c-a) cot B/2
= - k (cosA - cosB) + - k(cosB - cosC) + - k(cosC - cosA)
= - k(cosA - cosB + cosB - cosC + cosC - cosA)
= 0