Question

In a triangle, prove that (b+c-a)(cotB/2+cotC/2)=2a×cotA/2​

Answer 1

Answer:

How do we prove this? (a+b+c) ^2/ (a^2+b^2+c^2) = (cot(a/2) +cot(b/2) +cot(c/2)) / (cota+cotb+cotc)

(a+b+c)2a2+b2+c2=cota2+cotb2+cotc2cota+cotb+cotc(a+b+c)2a2+b2+c2=cot⁡a2+cot⁡b2+cot⁡c2cot⁡a+cot⁡b+cot⁡c

That can’t possibly be true can it? It’s an algebraic relation between transcendentally related quantities, unless there’s some deep cancelling going on.

It’s a little difficult to get a defined counterexample. The easiest I found is a=b=c=π/4a=b=c=π/4. We have cot(π/4)=1cot⁡(π/4)=1 of course and cot(π/8)=1+2–√cot⁡(π/8)=1+2; trust me. So the right side is

3(1+2–√)3=1+2–√3(1+2)3=1+2

Miraculously all the ππs vanish from the left and we’re left with

(3π/4)2/(3