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

= 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)