# Show $\tan(x)+\tan(y)+\tan(z) = \tan(x) \tan(y) \tan(z)$ [duplicate]

I am not able to show that:

If $x+y+z=\pi$, show that $\tan(x) + \tan(y) + \tan(z) = \tan(x) \tan(y) \tan(z)$.

• $x=y=\dfrac{\pi}{2},z=0$ will give you $x+y+z=\pi$ but $\tan(x)$ is not well-defined. – Hirshy Aug 13 '15 at 10:49
• You could explore $\tan (x_1+x_2+\dots x_r)$ - in terms of the $\tan x_i$. The result is a fraction with "odd" combinations of the $\tan x_i$ in the numerator and "even" combinations in the denominator. A little investigation will give you the rule for signs as well. Not hugely useful in general, but I find it a nice thing to know. And knowing $\tan (a+b+c)$ solves this easily. – Mark Bennet Aug 13 '15 at 11:29

## 3 Answers

A reasonable first step is to take the tangent of both sides of what you've been given; that gives $$\tan(x+y+z) = 0 \tag{\ast}$$ Now you have something involving the tangent of a sum of some numbers, and want something involving the tangents of the numbers themselves. Lucky for us, there are standard formulas for that. Usually you'll see one with two summands: $$\tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$$ On the left, tangent of sum; on the right, tangents of the summands. To apply this to ($\ast$), take, say, $a=x$ and $b=y+z$ and see where you end up.

Given $x+y+z = \pi\Rightarrow x+y = \pi-x$

Now taking $\tan$ on both side, we get $$\tan (x+y) = \tan(\pi-x) = -\tan x$$

So $$\displaystyle \frac{\tan x+\tan y}{1-\tan x\cdot \tan y} = -\tan x\Rightarrow \tan x+\tan y = -\tan z+\tan x\cdot \tan y\cdot \tan z$$

So we get $$\tan x+\tan y +\tan z = \tan x \cdot \tan y \cdot \tan z$$

• How do you get to $x+y=\pi-x$? – Hirshy Aug 13 '15 at 10:57
• Just a typo, but nice ! – Claude Leibovici Aug 13 '15 at 11:03
• @HasitBhatt $$\tan\left(\frac{\pi}{3}\right)=\sqrt{3}=-(-\sqrt{3})=-\tan\left(\frac{2\pi}{3}\right).$$ – Hirshy Aug 13 '15 at 11:10
• Is it allowed that if you have $x+y=\pi-z$, you can $\tan(x+y)=\tan(\pi - z)$? What rule, axiom, whatever states that this is allowed? Thanks for the answer. – reeglysonforubuntu Aug 13 '15 at 11:15

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

Now, $tan(x+y+z) = tan(\pi) = 0$, So,

$tan(x)+tan(y)+tan(z) - tan(x)tan(y)tan(z) = 0$.

$\implies tan(x)+tan(y)+tan(z) = tan(x)tan(y)tan(z)$