How to make use of 1 in helping us to solve International Math Olympiad problem?

Here is another blog post that describes how we could use the figure $1$ in solving intriguing International Olympiad Mathematical Problems.

According to wikipedia (number one (1)):

One, sometimes referred to as unity, is the integer before two and after zero. One is the first non-zero number in the natural numbers as well as the first odd number in the natural numbers.
Any number multiplied by one is that number, as one is the identity for multiplication. As a result, one is its own factorial, its own square, its own cube, and so on. One is also the result of the empty product, as any number multiplied by one is itself. It is also the only natural number that is neither composite nor prime with respect to division, but instead considered a unit.

The number "One" can be manipulated nicely to help us attacking some of the IMO problems successfully.

Note that

$\begin{align*}1&=1+0\\&=0+1\\&=1-x+x\\&=1+x-x\\&=(x+1)-x\end{align*}$

$\therefore 1^{\circ}=((x^{\circ}+1^{\circ})-x^{\circ})$

$\begin{align*}\sin 1^{\circ}&=\sin ((x^{\circ}+1^{\circ})-x^{\circ})\\&=\sin (x^{\circ}+1^{\circ})\cos x^{\circ}-\cos (x^{\circ}+1^{\circ})\sin x^{\circ}\end{align*}$

If we divide both sides of the equation by $\sin x \sin (x^{\circ}+1^{\circ})$, we get:

$\begin{align*}\dfrac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x^{\circ}+1^{\circ})}&=\dfrac{\sin (x^{\circ}+1^{\circ})\cos x^{\circ}-\cos (x^{\circ}+1^{\circ})\sin x^{\circ}}{\sin x^{\circ} \sin (x+1^{\circ})}\\&=\dfrac{\sin (x^{\circ}+1^{\circ})\cos x^{\circ}}{\sin x^{\circ} \sin (x+1^{\circ})}-\dfrac{\cos (x^{\circ}+1^{\circ})\sin x^{\circ}}{\sin x^{\circ} \sin (x^{\circ}+1^{\circ})}\\&=\dfrac{\cancel{\sin (x^{\circ}+1^{\circ})}\cos x^{\circ}}{\sin x^{\circ} \cancel{\sin (x^{\circ}+1^{\circ})}}-\dfrac{\cos (x^{\circ}+1^{\circ})\cancel{\sin x^{\circ}}}{\cancel{\sin x^{\circ}}\sin (x^{\circ}+1^{\circ})}\\&=\cot x^{\circ}-\cot (x^{\circ}+1^{\circ})\end{align*}$

Whenever we see that pattern, we know it would be a very useful identity in telescoping-related math series problems.

Here is one perfect problem that the trick above could be applied, it is a 21th USA Mathematical Olympiad 1992 Problem that I encourage you to tackle it before seeing the solution:

Let $k=1^{\circ}$, show that [MATH]\sum_{n=0}^{88}\dfrac{1}{\cos (nk) \cos(n+1)k}=\dfrac{\cos k}{\sin^2 k}[/MATH].

As usual, I will post the solution the next day, and if you have any comment, feel free to post it and I will take the time to reply to you.