Question

tan x + tan(x +pi/4) = 1.
help me fine x ..​

Answer 1

Recall that

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

and that $\tan\frac\pi4=1$. So we have

$\tan x+\tan\left(x+\dfrac\pi4\right)=1$

$\implies\tan x+\dfrac{\tan x+\tan\frac\pi4}{1-\tan x\tan\frac\pi4}=1$

$\implies\tan x+\dfrac{\tan x+1}{1-\tan x}=1$

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

$\implies\dfrac{1+2\tan x-\tan^2x}{1-\tan x}=1$

As long as $\tan x\neq1$, we can write

$\implies1+2\tan x-\tan^2x=1-\tan x$

$\implies3\tan x-\tan^2x=0$

$\implies\tan x(3-\tan x)=0$

$\implies\tan x=0\text{ or }\tan x=3$

$\implies\boxed{x=n\pi\text{ or }x=\tan^{-1}3+n\pi}$

where $n$ is any integer.