The value of $\tan \left(\frac{7\pi }{8}\right)$ is

Let

$A=\tan \left(\frac{7\pi }{8}\right)$

$\Rightarrow A=\tan \left(\mathrm{\pi }-\frac{\pi }{8}\right)$

$\Rightarrow A=-\tan \left(\frac{\pi }{8}\right)$

Now,

$\tan \left(\frac{\mathrm{\pi }}{4}\right)=\tan \left(2\times \frac{\mathrm{\pi }}{8}\right)=\frac{2\tan \left({\displaystyle \frac{\mathrm{\pi }}{8}}\right)}{1-{\tan }^2\left({\displaystyle \frac{\mathrm{\pi }}{8}}\right)}$

$\Rightarrow \frac{2\tan \left({\displaystyle \frac{\mathrm{\pi }}{8}}\right)}{1-{\tan }^2\left({\displaystyle \frac{\mathrm{\pi }}{8}}\right)}=1$

$\Rightarrow \frac{2\left(-A\right)}{1-{\left(-A\right)}^2}=1$

$\Rightarrow A^2-2A-1=0$

$\Rightarrow A=\frac{2\pm \sqrt{4+4}}{2}$

$\Rightarrow A=1\pm \sqrt{2}$

But $A<0$, since $\left(\frac{7\mathrm{\pi }}{8}\right)$ lies in $2^{\mathrm{nd}}$ quadrant, where tangent angle is negative, therefore

$A=1-\sqrt{2}$.