Formula of Tan(x+y+z) with Proof

The formula of tan(x+y+z) is given as follows:

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

In this post, let us learn how to prove the formula of tan(x+y+z).

Proof of tan(x+y+z) Formula

We will use the formula

tan(a+b) = $\dfrac{\tan a +\tan b}{1-\tan a \tan b}$.

Put a=x+y and b=z.

Now, tan(x+y+z) = tan((x+y)+z)

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

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

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

= $\dfrac{\tan x +\tan y+\tan z – \tan x \tan y \tan z}{1-\tan x \tan y-\tan y \tan z-\tan z \tan x}$.

So the formula of tan(x+y+z) is equal to (tanx + tany +tanz – tanx tany tanz)/(1- tan x tany – tany tanz – tan z tanx).

More Formulas:

Formula of sin(x+y+z)

Formula of cos(x+y+z)

FAQs