Prove $\tan 54^\circ=\frac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$

How to prove this identity without using the actual values of $\tan54^\circ$ and $\sin24^\circ$

$$\tan 54^\circ=\dfrac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$$

Edit: I still don’t get it, I am stuck at:

$$\dfrac{\cos 24^\circ+\sqrt{3}\sin 24^\circ}{\sqrt{3}\cos 24^\circ-\sin 24^\circ}$$

P.S. : Is this coincidence or a more general form can be obtained?

Solutions Collecting From Web of "Prove $\tan 54^\circ=\frac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$"

Finally,,,after much effort I have a proof.

Note that $2\cdot 36^{\circ} =180^{\circ}-3\cdot 36^{\circ}$ therefore

$$\sin 2\cdot 36^{\circ}\cdot =\sin 3\cdot 36^{\circ}$$ and thus

$$2\sin36^{\circ}\cos 36^{\circ}=3\sin 36^{\circ}-4\sin^336^{\circ}$$
and therefore

$$2\cos 36^{\circ}=3-4\sin^2 36^{\circ}$$
so
$$2\cos 36^{\circ}=3\cos^2 36^{\circ}-\sin^2 36^{\circ}=(\sqrt{3}\cos36^{\circ}-\sin 36^{\circ})(\sqrt{3}\cos36^{\circ}+\sin36^{\circ})$$

Now
$$\sin24^{\circ}=\sin(60^{\circ}-36^{\circ})=\frac{\sqrt{3}}{2}\cos36^{\circ}-\frac{1}{2}\sin36^{\circ}$$

and therefore with the last equation we have
$$\cos 36^{\circ}=\sin24^{\circ}(\sqrt{3}\cos36^{\circ}+\sin36^{\circ})$$

dividing by $\sin36^{\circ}$ we get

$$\cot 36^{\circ}=\sin24^{\circ}(\sqrt{3}\cot36^{\circ}+1)$$

Note that $\tan54^{\circ}=\tan(90^{\circ}-36^{\circ})=\cot36^{\circ}$

$$\tan54^{\circ}=\sin24^{\circ}(\sqrt{3}\tan54^{\circ}+1)$$ from which

$$\tan54^{\circ}=\frac{\sin24^{\circ}}{1-\sqrt{3}\sin24^{\circ}}$$
follows.

The moral is that $2\cdot 18^{\circ}+3\cdot 18^{\circ}=90^{\circ}$ yields some nice trig relations that should be taken note of.

Hint:

$$\tan 54^\circ = \tan (30^\circ + 24^\circ) = \frac{\sin(30^\circ+24^\circ)}{\cos(30^\circ+24^\circ)}$$