louisatanhuiying
Banned
- Joined
- Sep 24, 2012
- Messages
- 6
If A, B and C are the angles of a triangle, prove:
tan (A/2) tan (B/2) + tan (B/2) tan (C/2) + tan (C/2) tan (A/2) = 1
I'm guessing since A + B + C = 180o, (a) should be solved with tan 90o. 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)] / (tan90o) which can't be solved because nothing can legitimately be divided by tan90o.
Sorry if you've seen this question under the thread of 'trigo surds question', it's meant to be here! And if you've been working on part (b), I removed it because I figured it out. Now I'm still working on (a).Thanks!
tan (A/2) tan (B/2) + tan (B/2) tan (C/2) + tan (C/2) tan (A/2) = 1
I'm guessing since A + B + C = 180o, (a) should be solved with tan 90o. 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)] / (tan90o) which can't be solved because nothing can legitimately be divided by tan90o.
Sorry if you've seen this question under the thread of 'trigo surds question', it's meant to be here! And if you've been working on part (b), I removed it because I figured it out. Now I'm still working on (a).Thanks!
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