Tangents

**TheodorMunteanu** · July 17th 2011, 01:22 PM

if $x+y+z+t=\pi$ then $\tan x+\tan y+\tan z+\tan t=\tan x\tan y\tan z\tan t$

**Also sprach Zarathustra** · July 17th 2011, 01:54 PM

Originally Posted by TheodorMunteanu

if $x+y+z+t=\pi$ then $\tan x+\tan y+\tan z+\tan t=\tan x\tan y\tan z\tan t$

Hint:

$\tan x+\tan y+\tan z+\tan t=$

$=\frac{1}{4}\sec t\sec x\sec y\sec z \cdot(-\sin(t-x-y-z)+\sin(t+x+y-z)+\sin(z+x-y+z)+\sin(t-x+y+z)+2\sin(x+y+z+t))$

**TheodorMunteanu** · July 17th 2011, 02:46 PM

Actually you're trying to prove a wrong exercise:
Correct it would have been: $\tan x+\tan y+\tan z+\tan t=\frac{\tan x+\tan y+\tan z+\tan t-\tan x\tan y\tan z-tan x\tan z \tan t-\tan x\tan y\tan t-\tan y\tan z\tan t}{\tan x\tan y+...-\tan x\tan y\tan z\tan t}$

**Soroban** · July 17th 2011, 04:32 PM

Hello, TheodorMunteanu!

$\text{if }x+y+z+t\,=\,\pi,$

$\text{ then: }\:\tan x+\tan y+\tan z+\tan t\:=\:\tan x\tan y\tan z\tan t$

This is not true!
$\text{Let }x = y = z = t = \tfrac{\pi}{4}$

$\text{We have: }\:x + y + z + t \:=\:\pi$

$\text{Take tangents: }\:\tan(x+y + z + t) \;=\;\tan(0)$

. . . . . . . . $\tan\bigg[(x+y) + (z+t)\bigg] \;=\;0$

. . . . . . . $\frac{\tan(x+y) + \tan(z+t)}{1 - \tan(x+y)\tan(z+t)} \;=\;0$

Assume that the denominator is not zero.
A fraction equals zero if its numerator equals zero.

$\text{We have: }\:\tan(x+y) + \tan(z+t) \;=\;0$

. . $\frac{\tan x + \tan y}{1 - \tan x\tan y} + \frac{\tan z + \tan t}{1 - \tan z\tan t}\;=\;0$

$\text{Then: }\,\frac{(\tan\!x +\tan\!y)(1-\tan\!z\tan\!t) + (\tan\!z + \tan\!t)(1 - \tan\!x\tan\!y)}{(1-\tan x\tan y)(1 - \tan z\tan t)} \;=\;0$

Assume that the denominator is not zero.
A fraction equals zero if its numerator equals zero.

$(\tan\!x +\tan\!y)(1-\tan\!z\tan\!t) + (\tan\!z + \tan\!t)(1 - \tan\!x\tan\!y) \;=\;0$

$\tan x - \tan x\tan z\tan t + \tan y - \tan y \tan z\tan t$
. . . $+ \tan z - \tan x \tan y\tan z + \tan t - \tan x\tan y \tan t \;=\;0$

Therefore:

$\tan x + \tan y + \tan z + \tan t$

. . $=\;\tan x\tan y\tan z + \tan x\tan y\tan t + \tan x\tan z\tan t + \tan y\tan z\tan t$

**TheodorMunteanu** · July 18th 2011, 10:37 AM

Yeah well as you can see I've corrected my previous post 2 hours earlier.

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