Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Well, at first, the problem has two typos (one from the original source and one that occured during the LaTeXing). Here is the correct formulation: Problem. Prove the identity $\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\dfrac{x}{2}\right)^{2k}\left(1+\dfrac{2^k}{\left(1-\tan^2\dfrac{x}{2}\right)^k}\right)=\sec^{2n}\dfrac{x}{2}+\sec^n x$ for any natural number n and any angle x. For the solution, denote $A=\tan^2\dfrac{x}{2}$. Then, $A+1=\tan^2\dfrac{x}{2}+1=\dfrac{\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}+1=\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}=\dfrac{1}{\cos^2\dfrac{x}{2}}=\sec^2\dfrac{x}{2}$ and $\dfrac{1+A}{1-A}=\dfrac{1+\tan^2\dfrac{x}{2}}{1-\tan^2\dfrac{x}{2}}=\dfrac{1+\dfrac{\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}}{1-\dfrac{\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}}=\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}=\dfrac{1}{\cos x}=\sec x$. Thus, $\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\dfrac{x}{2}\right)^{2k}\left(1+\dfrac{2^k}{\left(1-\tan^2\dfrac{x}{2}\right)^k}\right)=\sum\limits_{k=0}^n\binom{n}{k}\left(\tan^2\dfrac{x}{2}\right)^{k}\left(1+\dfrac{2^k}{\left(1-\tan^2\dfrac{x}{2}\right)^k}\right)$ $=\sum\limits_{k=0}^n\binom{n}{k}A^k\left(1+\dfrac{2^k}{\left(1-A\right)^k}\right)=\sum\limits_{k=0}^n\binom{n}{k}A^k+\sum\limits_{k=0}^n\binom{n}{k}A^k\dfrac{2^k}{\left(1-A\right)^k}$ $=\sum\limits_{k=0}^n\binom{n}{k}A^k+\sum\limits_{k=0}^n\binom{n}{k}\left(\dfrac{2A}{1-A}\right)^k=\left(A+1\right)^n+\left(\dfrac{2A}{1-A}+1\right)^n$ $=\left(A+1\right)^n+\left(\dfrac{1+A}{1-A}\right)^n=\left(\sec^2\dfrac{x}{2}\right)^n+\left(\sec x\right)^n=\sec^{2n}\dfrac{x}{2}+\sec^n x$, and the problem is solved. darij