(i) Show that, if $\tan^2(x) = 2\tan(x) + 1,$ then $\tan (2x) = -1$
The desired result suggests playing with the double-angle formulas: $$\sin(2x) = 2\sin(x)\cos(x)$$ $$\cos(2x) = 2\cos^2(x) - 1$$
Also, since you're given a square tangent, you may as well try using a pythagorean identity $\tan^2 + 1 = \sec^2$.
Now, $$\tan^2(x) = 2\tan(x) + 1 = \sec^2(x) - 1$$ implies $$\sec^2(x) = 2\tan(x) + 2 = 2(\tan(x) + 1)$$ That is, $$1 = 2(\tan(x) + 1)\cos^2(x) = 2\sin(x)\cos(x) + 2\cos^2(x)$$ Pull the one over to the right side. Do you see how to finish it off?