Originally Posted by
David24 hey mate,
I'll take on the challenge!
As I'm sure you know cot(x) = cos(x)/sin(x)
Thus for each A,B,C we need to resolve cot(x) or moreover sin(x), cos(x)
for cot(A)
(c^2 - (a^2 + b^2))/(2ac*sin(B))
Cosine Rule :
a^2 = b^2 + c^2 - 2bc*cos(A)
Thus cos(A) = (a^2 - (b^2 + c^2)) / (2bc)
Sine Rule :
a/sin(A) = b/sin(B)
Thus,
sin(A) =(a/b)sin(B)
Hence,
cot(A) = (((b^2 + c^2)-a^2) / 2bc)/ ((a/b)sin(B))
cot(A) = (( (b^2 + c^2)-a^2) / (2ac * sin(B)))
for cot(B)
Cosine Rule :
b^2 = a^2 + c^2 - 2ac*cos(B)
cos(B) = ((a^2 + c^2)-b^2) /(2ac)
Thus,
cot(B) = (((a^2 + c^2)-b^2) /(2ac))/sin(B)
cot(B) = (((b^2 + c^2)-b^2) / (2ac * sin(B)))
for cot(C)
Cosine Rule :
c^2 = a^2 + b^2 - 2ab*cos(C)
cos(C) = (c^2 - (a^2 + b^2))/(2ab)
Sine Rule:
b/sin(B) = c/sin(C)
Thus,
sin(C) = (c/b)sin(B)
Thus,
cot(C) = ((c^2 - (a^2 + b^2))/(2ab))/ ( (c/b)sin(B) )
cot(C) = (c^2 - (a^2 + b^2))/(2ac*sin(B))
Hence we now can define cot(A) + cot(B) + cot(C)
cot(A) + cot(B) + cot(C)
= (( (b^2 + c^2)-a^2) / (2ac * sin(B))) + cot(B) = (((a^2 + c^2)-b^2) / (2ac * sin(B))) + ((a^2 + b^2)-c^2)/(2ac*sin(B))
= (1/(2ac * sin(B))(a^2 + b^2 + c^2)
Area of ABC = (1/2)(ac * sin(B) --> ac*sin(B) = 2 *(Area of ABC)
Sub in,
cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/ (2 * (2 * (Area of ABC)))
cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/(4 x (Area of ABC))
= RHS
Thus true,
Hope this helps,
David