Recall the half angle identities for sine and cosine:
\(\sin^2x=\dfrac{1-\cos2x}2\)
\(\cos^2x=\dfrac{1+\cos2x}2\)
So you have
\(\tan^2x=\dfrac{1-\cos2x}{1+\cos2x}\)
This is useful because we have
\(\tan^2\dfrac\pi{24}=\dfrac{1-\cos\dfrac\pi{12}}{1+\cos\dfrac\pi{12}}\)
Unfortunately, we need to know the value of \(\cos\dfrac\pi{12}\) to continue. But we can use either of the half-angle identities to figure this out.
\(\cos^2\dfrac\pi{12}=\dfrac{1+\cos\dfrac\pi6}2\)
\(\cos^2\dfrac\pi{12}=\dfrac{1+\dfrac{\sqrt3}2}2\)
\(\cos^2\dfrac\pi{12}=\dfrac{2+\sqrt3}4\)
\(\cos\dfrac\pi{12}=\dfrac{\sqrt{2+\sqrt3}}2\)
where you take the positive root since \(\cos x\) is positive when \(0\le x<\dfrac\pi2\).
So, you have
\(\tan^2\dfrac\pi{24}=\dfrac{1-\dfrac{\sqrt{2+\sqrt3}}2}{1+\dfrac{\sqrt{2+\sqrt3}}2}\)
\(\tan^2\dfrac\pi{24}=\dfrac{2-\sqrt{2+\sqrt3}}{2+\sqrt{2+\sqrt3}}\)
\(\tan\dfrac\pi{24}=\sqrt{\dfrac{2-\sqrt{2+\sqrt3}}{2+\sqrt{2+\sqrt3}}}\)
again taking the positive root because \(\tan x\) is positive when \(0.
