\(\tan^2x=\dfrac{\sin^2x}{\cos^2x}\)
\(\tan^2x=\dfrac{\frac{1-\cos2x}2}{\frac{1+\cos2x}2}\)
\(\tan^2x=\dfrac{1-\cos2x}{1+\cos2x}\)
\(\tan x=\pm\sqrt{\dfrac{1-\cos2x}{1+\cos2x}}\)
When \(\dfrac\pi2, you have \(\tan x<0\), so for \(x=\dfrac{7\pi}8\) you would take the negative root. Now,
\(\tan\dfrac{7\pi}8=-\sqrt{\dfrac{1-\cos\frac{7\pi}4}{1+\cos\frac{7\pi}4}}\)
\(\tan\dfrac{7\pi}8=-\sqrt{\dfrac{1-\frac1{\sqrt2}}{1+\frac1{\sqrt2}}}\)
\(\tan\dfrac{7\pi}8=-\sqrt{3-2\sqrt2}\)
