Question

in triangle abc prove that B minus C cot a by 2 + c minus A cot B by 2 + a minus b cos C by 2 is equal to zero​

Answer 1

Answer:

(b-c)Cot(A/2) + (c-a)Cot(B/2) + (a -b)Cot(C/2) = 0

Step-by-step explanation:

(b-c)Cot(A/2) + (c-a)Cot(B/2) + (a -b)Cot(C/2) = 0

a/SinA  = bSinB = c/SinC = k

=> a = kSinA  , b = kSinB  , c = kSinC

LHS

= k { (SinB - SInC)Cot(A/2)  + (SinC - SinA)Cot(B/2) + (SinA - SinB)Cot(C/2)}

Sinx - Siny = 2Sin((x - y)/2)Cos((x+y)/2)

= k { 2Sin((B-C)/2)Cos((B+C)/2)Cot(A/2) + 2Sin((C-A)/2)Cos((C+A)/2)Cot(B/2) + 2Sin((A-B)/2)Cos((A+B)/2)Cot(C/2) }

A + B + C = π

=> A + B = π- C

=> (A + B)/2 = π/2 - C/2

=> Cos (A + B)/2 = Cos( π/2 - C/2) = Sin(C/2)

Simiallrly Cos(B + C)/2  = Sin(A/2)   & Cos(C + A)/2 = Sin(B/2)

= k { 2Sin((B-C)/2)Sin(A/2)Cot(A/2) + 2Sin((C-A)/2)Sin(B/2)Cot(B/2) + 2Sin((A-B)/2)Sin(C/2)Cot(C/2) }

Cotx Sinx = Cosx

=  k { 2Sin((B-C)/2)Cos(A/2) + 2Sin((C-A)/2)Cos(B/2) + 2Sin((A-B)/2)Cos(C/2) }

= k { 2Sin((B-C)/2)Sin((B+C)/2) + 2Sin((C-A)/2)Sin((C+A)/2) + 2Sin((A-B)/2)Sin((A+B)/2) }

Using Sin(x + y)Sin(x - y) = Sin²x - Sin²y

= 2k ( Sin²B - Sin²C  + Sin²C - Sin²A + Sin²A - Sin²B)

= 2k * 0

= 0

= RHS

QED

Proved

(b-c)Cot(A/2) + (c-a)Cot(B/2) + (a -b)Cot(C/2) = 0