Question

In triangle ABC, prove that`tan(B-C)/2=(b-c)/(b+c)cotA/2``tan(C-A)/2=(c-a)/(c+a)cotB/2``tan(A-B)/2=(a-b)/(a+b)cotC/2`

Answer 1

From sine law, we have, `a/sinA = b/sinB = c/sinC = k`
`:.a = ksinA, b = ksinB, c = ksinC`
`:.(b-c)/(b+c) = (k(sinB-sinC))/(k(sinB+sinC))`
`= (sinB - sinC)/(sinB+SinC)`
`= (2cos((B+C)/2)sin((B-C)/2))/(2sin((B+C)/2)cos((B-C)/2))`
` = cot((B+C)/2)tan((B-C)/2)`
Here, `A+B+C = 180`
So, `B+C = 180-A`
`:.(b-c)/(b+c) = cot((180-A)/2)tan((B-C)/2) = cot(90-A/2)tan((B-C)/2) `
`(b-c)/(b+c) = tan(A/2)tan((B-C)/2)`
`tan((B-C)/2) = (b-c)/(b+c) cot(A/2)`
Similarly, we can show that,
`tan((C-A)/2) = (c-a)/(c+a) cot(B/2)`
`tan((A-B)/2) = (a-b)/(a+b) cot(C/2)`