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

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September 23

Mystery trig identity

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

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)

However, if A+B+C = 180 degrees, we do have

${\frac {1}{\tan A\tan B}}+{\frac {1}{\tan B\tan C}}+{\frac {1}{\tan C\tan A}}=1$

so maybe that is the real question. Gandalf61 (talk) 14:33, 23 September 2010 (UTC)

From the standard identity

$\tan(x+y)={\frac {\tan x+\tan y}{1-\tan x\tan y}}$

we can see that

$\tan(x+y+z)={\frac {\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan x\tan z-\tan y\tan z}}$

and then consider what happens when

$x+y+z={\text{half-circle}}=180^{\circ }.\,$

What happens is that in that case,

$\tan(z+y+z)=0,\,$

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

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

$x+y+z=90^{\circ }\,$

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