Prove that $\cot 99^\circ=3\cot36^\circ-\sec18^\circ-\csc18^\circ$

Question

Prove that:

$$\cot 99^\circ=3\cot36^\circ-\sec18^\circ-\csc18^\circ$$

Actually, I have a proof here (see edit section at the bottom of the text), which I needed to solve a geometry problem fully. It's definitely correct but I don't like it. The proof seems to be simple and short but it's very hard to do it completely by hand without some machine help. So I'm looking for a more ellegant proof and also hoping to learn something new.

Answer 1

As $\cot(90^\circ+y)=-\tan y,$

$$\dfrac1{\sin2x}-\tan x=\dfrac{1-2\sin^2x}{\sin2x}=\cot2x$$

We need

$$\cot2x+\dfrac1{\cos2x}=\dfrac{3\cos4x}{\sin4x}$$

$$\iff2\cos^22x+2\sin2x=3\cos4x$$

If $s=\sin2x,$

$$\iff2(1-s^2)+2s=3(1-2s^2)$$

$$4s^2+2s-1=0$$

So, $s=\sin18^\circ$ or $=-\sin54^\circ$

Can you find all the possible values of $x$ including $9^\circ$

See Various methods to find value of $\sin 18^\circ$