Remember the half angle identity for the tangent:
[tex]\tan (\frac{x}{2})=\frac{\sin x}{1+\cos x}[/tex]Since 7/8 = (7/4)/2, then:
[tex]\tan (\frac{7\pi}{8})=\tan (\frac{7\pi}{4}/2)=\frac{\sin (\frac{7\pi}{4})}{1+\cos (\frac{7\pi}{4})}[/tex]The values of sin(7/4) and cos(7/4) are well known:
[tex]\begin{gathered} \sin \mleft(\frac{7}{4}\mright)=-\frac{\sqrt[]{2}}{2} \\ \cos \mleft(\frac{7\pi}{4}\mright)=\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]Then:
[tex]\begin{gathered} \tan \mleft(\frac{7\pi}{8}\mright)=\frac{-\frac{\sqrt[]{2}}{2}}{1+\frac{\sqrt[]{2}}{2}} \\ =-\frac{\sqrt[]{2}}{2+\sqrt[]{2}} \\ =-\frac{\sqrt[]{2}}{2+\sqrt[]{2}}\cdot\frac{2-\sqrt[]{2}}{2-\sqrt[]{2}} \\ =-\frac{\sqrt[]{2}\times(2-\sqrt[]{2})}{4-2} \\ =-\frac{2\sqrt[]{2}-2}{2} \\ =\frac{2-2\sqrt[]{2}}{2} \\ =1-\sqrt[]{2} \end{gathered}[/tex]Therefore:
[tex]\tan (\frac{7\pi}{8})=1-\sqrt[]{2}[/tex]