Slovenian Mathematical Olympiad, 2008

From the addition formula for tangents we get
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\tan \frac{x}{2} + \tan \frac{\pi}{4}}{1 - \tan \frac{x}{2} \tan \frac{\pi}{4}} \right)^2 .$$
Using the fact that $\tan \frac{\pi}{4} = 1$ and expressing $\tan \frac{x}{2}$ in terms of sines and cosines, we see that
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} + 1}{1 - \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}} \right)^2 = \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2 .$$
We now square both sides of the equation to get
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{\cos^2 \frac{x}{2} + 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} - 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}$$
and use the relations $\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} = 1$ and $2 \cos \frac{x}{2} \sin \frac{x}{2} = \sin x$ to show that
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{1 + \sin x}{1 - \sin x} = \frac{2}{1 - \sin x} - 1.$$
Thus, the equality holds.

Solution 2:
Express the tangent in terms of sine and cosine
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\sin \left( \frac{x}{2} + \frac{\pi}{4} \right)}{\cos \left( \frac{x}{2} + \frac{\pi}{4} \right)} \right)^2 .$$
Now, use the addition formulas for sine and cosine to show that
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\sin \frac{x}{2} \cos \frac{\pi}{4} + \cos \frac{x}{2} \sin \frac{\pi}{4}}{\cos \frac{x}{2} \cos \frac{\pi}{4} - \sin \frac{x}{2} \sin \frac{\pi}{4}} \right)^2 .$$
We know that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so the expression is equal to
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\frac{\sqrt{2}}{2} \sin \frac{x}{2} + \frac{\sqrt{2}}{2} \cos \frac{x}{2}}{\frac{\sqrt{2}}{2} \cos \frac{x}{2} - \frac{\sqrt{2}}{2} \sin \frac{x}{2}} \right)^2 .$$
Squaring both sides we get
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{\cos^2 \frac{x}{2} + 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} - 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}$$
and the equalities $\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} = 1$ and $2 \cos \frac{x}{2} \sin \frac{x}{2} = \sin x$ imply
$$\tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{1 + \sin x}{1 - \sin x} = \frac{2}{1 - \sin x} - 1.$$