
Prove that

Quote:
Originally Posted by
matsci0000 c(acosBbcosA)=a^2b^2
Can we have the rest of the question please, is this a triangle with sides a,b,c and angles opposite the sides A, B, C?
Try the difference of the cosine rule for a and b.
CB

Quote:
Originally Posted by
matsci0000 c(acosBbcosA)=a^2b^2
According to cosine rule
a^2 = b^2 + c^2  2bc*cosA.
From this find bc*cosA. Similarly find ac*coaB, And find the difference.
Or, by using projection rule write c as
c = acosB + bcosA and simplify.