21th USA Mathematical Olympiad 1992 Problem

In the (previous blog post), we were asked to prove a 21th USA Mathematical Olympiad 1992 Problem:

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]

But we have already worked out an trigonometric identity where:

$\dfrac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x^{\circ}+1^{\circ})}=\cot x^{\circ}-\cot (x^{\circ}+1^{\circ})$

Analysis Quiz 7: IMO Mock Trigonometric Math Quiz

Analysis Quiz 6:

IMO Mock Trigonometric Math Quiz

Please answer the following questions based on the trigonometric equation below:

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

Question 1.

Would you convert the cosecant function to the sine function to solve the trigonometric equation above?

Yes.
No.
Perhaps.

Answer:

For an old hand like me (I'm not old at all, hehehe...), I could say right up front, loudly that I would not convert the cosecant function to the sine function in this case! But, if you are new and eager to learn, I will lead you to the answer, just that you have to be patient.

Quiz 7: IMO Mock Trigonometric Math Quiz

Alternative Way to Prove $\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$

So in one of my recent post, I mentioned of one rare but extremely useful trigonometry identity that sounds:

$\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$

We should then use it when appropriate.