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The expression I was trying to simplify was this :

$$\frac {\sin x} {\cos 3x}+ \frac{\sin 3x}{\cos 9x} +\frac{\sin 9x}{\cos 27 x}$$

This question does have answers here Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$

And also here and here : https://math.stackexchange.com/a/1299968

Concept of Trigonometric identities

However, all of the answers in these questions are the same method; the method that I have given below.

I was working with this expression for a long time, and finally, I looked it up online.

The only way I saw online to simplify this is by:- $$\frac {\sin x} {\cos 3x}=\frac {\sin x} {\cos 3x} * \frac {\cos x} {\cos x}$$ $$= \frac {\sin x \cos x} {\cos 3x\cos x}$$ $$= \frac{\sin 2x}{2*\cos 3x \cos x}$$ $$=\frac 1 2 *\frac {\sin 2x} {\cos 3x \cos x}$$ $$=\frac 1 2 *\frac{\sin (3x-x)}{\cos 3x \cos x}$$ $$=\frac 1 2 * \frac {\sin 3x\cos x -\cos 3x\sin x} {\cos 3x\cos x}$$ $$= \frac 1 2 * (\tan 3x - \tan x)$$

So you first have to know to multiply this expression with $\frac {\cos x} {\cos x}$, then you have to realise that $\sin x\cos x = \frac {\sin 2x} 2$. and then you have to realise that $\sin 2x= \sin (3x-x)$.

This is very unintuitive. The first thing anyone not familiar with this would try to do is to convert $\cos 3x$ in terms of $\cos x$, and then work from there.

So this is my question. Is there any intuitive way to work out this expression?

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