Trigo surds question

[louisatanhuiying]

Trigo angle qn

If A, B and C are the angles of a triangle, prove:

(a) tan (A/2) tan (B/2) + tan (B/2) tan (C/2) + tan (C/2) tan (A/2) = 1
(b) sin A + sin B + sin C = 4 cos (A/2) cos (B/2) cos (C/2)

I'm guessing since A + B + C = 180^o, (a) should be solved with tan 90^o. However, the answer I've amounted to is always the right hand side = 1 + [tan(A/2) × tan(B/2) × tan(C/2) - tan(A/2) - tan(B/2) - tan(C/2)] / (tan90^o) which can't be solved because nothing can legitimately be divided by tan90^o. Meanwhile, for the right hand side of (b), I've always been stuck at 4 cos (A/2) cos (B/2) cos (C/2) + 2[cos (A/2) cos (B/2) cos (C/2) - cos (A/2) sin (B/2) sin (C/2) - sin (A/2) cos (B/2) sin (C/2) - sin (A/2) sin (B/2) cos (C/2)]. Thank you!

 

[tkhunny]

Two Hints that may lead to something:

1) 3 + 2 = 5

2) 5*36 = 180