Slovenija 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 $${\tan }^2\left(\frac{x}{2}+\frac{\pi }{4}\right)=\frac{1+\sin x}{1-\sin x}=\frac{2}{1-\sin x}-1.$$