In triangle ABC, prove that:

tan((B − C)/2) = (b − c)/(b + c) × cot(A/2)

tan((C − A)/2) = (c − a)/(c + a) × cot(B/2)

tan((A − B)/2) = (a − b)/(a + b) × cot(C/2)

In triangle ABC, prove that

tan((B − C)/2) = (b − c)/(b + c) × cot(A/2)

tan((C − A)/2) = (c − a)/(c + a) × cot(B/2)

tan((A − B)/2) = (a − b)/(a + b) × cot(C/2)

Proof by sine formula, we have

a/sin(A) = b/sin(B) = c/sin(C) = k (say)

Therefore, (b − c)/(b + c) = k(sin(B) − sin(C)) / k(sin(B) + sin(C))

= [2cos((B + C)/2) × sin((B − C)/2)] / [2sin((B + C)/2) × cos((B − C)/2)]

= cot((B + C)/2) × tan((B − C)/2)

= cot(π/2 − A/2) × tan((B − C)/2)

⇒ tan((B − C)/2) = (b − c)/(b + c) × cot(A/2)

Therefore, tan((B − C)/2) = (b − c)/(b + c) × cot(A/2)

Similarly, (ii) and (iii) can be proved.