tan A/2 tan B/2 + tan B/2 tan C/2 +tan C/2 tan A/2 = 1

Question

please help me

Answer 1

This cannot be an identity for all arbitrary values of A, B and C, because, for example, if A=B=C=0 then 0=1 which is false, so we need to assume a connection between A, B and C. Let's assume A+B+C=180° (as in a triangle). Then C=180-(A+B).

Now let A=B=C=60 (equilateral triangle) and check: 

3tan(30)tan(30)=3(1√3)²=1. So the identity is true in this case.

Let A=B=45°, so C=90° and tan(½C)=1.

tan(2x)=2tan(x)/(1-tan²(x)) so if x=22.5°, then 1=2tan(22.5)/(1-tan²(22.5)). Let t= tan(22.5), then:

1=2t/(1-t²), 1-t²=2t, t²+2t=1, t²+2t+1=2, (t+1)²=2, t+1=±√2, t=√2-1 gives us a positive angle, so tan(22.5)=√2-1.

Put the proposed identity to the test:

tan²(22.5)+tan(22.5)+tan(22.5)=(√2-1)²+√2-1+√2-1=

3-2√2+2√2-2=1, and the identity is true.

Now we can be reasonably sure that if C=180-(A+B), the identity is always true.

If C=180-(A+B), ½C=90-½(A+B). tan(½C)=cot(½(A+B))=1/tan(½(A+B)).

tan(½A+½B)=(tan(½A)+tan(½B))/(1-tan(½A)tan(½B));

tan(½C)=1/tan(½(A+B))=(1-tan(½A)tan(½B))/(tan(½A)+tan(½B))

So the LHS of the proposed identity is:

tan(½A)tan(½B)+tan(½C)(tan(½A)+tan(½B))=

tan(½A)tan(½B)+(1-tan(½A)tan(½B))[(tan(½A)+tan(½B))/(tan(½A)+tan(½B))]=

tan(½A)tan(½B)+1-tan(½A)tan(½B)=1 QED