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Wikipedia:Reference desk/Archives/Mathematics/2010 September 23

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September 23

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how to prove the question M tanAtanB+tanBtanC+tanCtanA=1, if A+B+C=180° —Preceding unsigned comment added by Siddhiraj Khanal (talkcontribs) 11:38, 23 September 2010 (UTC)[reply]

What is M? Note that in general, tanAtanB+tanBtanC+tanCtanA≠1. Consider an equilateral triangle; tan60°=√3, so the sum would be 9. Your formula does no appear to be related to the Law of tangents. I added a section title.-- 114.128.149.193 (talk) 14:10, 23 September 2010 (UTC)[reply]
However, if A+B+C = 180 degrees, we do have
so maybe that is the real question. Gandalf61 (talk) 14:33, 23 September 2010 (UTC)[reply]

From the standard identity

we can see that

and then consider what happens when

What happens is that in that case,

so the numerator on the other side of the fraction must be zero. Michael Hardy (talk) 20:21, 23 September 2010 (UTC)[reply]

....Oh.....that's not quite what you were asking. What you need is for the denominator to be zero. That happens if

And what is your "M"? Michael Hardy (talk) 20:23, 23 September 2010 (UTC)[reply]