In triangle A B C ,   a + b + c B C + A B + a + b + c A C + A B = 3 then tan C 8 =

Given, $\frac{a+b+c}{BC+AB}+\frac{a+b+c}{AC+AB}=3$

Here, $BC=a, AC=b, AB=c$.

Then, $\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$

$\Rightarrow \frac{a+c}{a+c}+\frac{b}{a+c}+\frac{b+c}{b+c}+\frac{a}{b+c}=3$

$\Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3$

$\Rightarrow \frac{b}{a+c}+\frac{a}{b+c}=1$

$\Rightarrow b^2+bc+a^2+ac=ab+ac+bc+c^2$

$\Rightarrow \frac{b^2+a^2-c^2}{ab}=1$

$\Rightarrow \frac{b^2+a^2-c^2}{2ab}=\frac{1}{2}$

Now from Cosine rule, we know $\frac{b^2+a^2-c^2}{2ab}=\cos C$

$\Rightarrow \cos C=\frac{1}{2}$

$\Rightarrow C=\frac{\pi }{3}$

Now, $\tan \frac{C}{8}=\tan \left(\frac{\pi }{3\times 8}\right)=\tan \left(\frac{\pi }{24}\right)$

Let $\frac{\pi }{24}=x\Rightarrow \frac{\pi }{6}=4x$

We know that $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan 4x=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{2\tan 2x}{1-{\tan }^{2}2x}=\frac{1}{\sqrt{3}}$

$\Rightarrow 2\sqrt{3}\tan 2x=1-{\tan }^{2}2x$

$\Rightarrow {\tan }^{2}2x+2\sqrt{3}\tan 2x-1=0$

$\Rightarrow \tan 2x=\frac{-2\sqrt{3}+\sqrt{12+4}}{2}=2-\sqrt{3}$

$\Rightarrow \frac{2\tan x}{1-{\tan }^{2}x}=2-\sqrt{3}$

$\Rightarrow \left(2-\sqrt{3}\right)\left({\tan }^{2}x\right)+2\tan x-\left(2-\sqrt{3}\right)=0$

$\Rightarrow \tan x=\frac{-2\pm \sqrt{4+4\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2\left(2-\sqrt{3}\right)}=\frac{-2\pm 2\sqrt{8-4\sqrt{3}}}{2\left(2-\sqrt{3}\right)}$

$\Rightarrow \tan x=\frac{-1\pm 2\sqrt{2-\sqrt{3}}}{\left(2-\sqrt{3}\right)}$

$\Rightarrow \tan x=\frac{-1\pm \sqrt{2}\left(\sqrt{3}-1\right)}{\left(2-\sqrt{3}\right)}$

Considering $\tan x=\frac{-1+\sqrt{2}\left(\sqrt{3}-1\right)}{\left(2-\sqrt{3}\right)}$

$\Rightarrow \tan x=\left(\sqrt{6}-\sqrt{2}-1\right)\left(2+\sqrt{3}\right)$

$\Rightarrow \tan x=\sqrt{6}+\sqrt{2}-2-\sqrt{3}$

Hence, $\tan \frac{C}{8}=\tan \frac{\pi }{24}=\sqrt{6}+\sqrt{2}-2-\sqrt{3}$.