Absurd Trignometric angles

Here I will use the product to sum formula, then the sum to product formula, also the fact that $2\sin(30^\circ)=1$ and $2\cos(30^\circ)=\sqrt{3}$ :

$4\sin(10^\circ)+\dfrac{\sqrt{3}\sin(10^\circ)}{\cos(10^\circ)}$

$\dfrac{4\sin(10^\circ)\cos(10^\circ)+\sqrt{3}\sin(10^\circ)}{\cos(10^\circ)}$

$\dfrac{2\sin(20^\circ)+2\sin(10^\circ)\cos(30^\circ)}{\cos(10^\circ)}$

$\dfrac{2\sin(20^\circ)+\sin(40^\circ)-\sin(20^\circ)}{\cos(10^\circ)}$

$\dfrac{\sin(20^\circ)+\sin(40^\circ)}{\cos(10^\circ)}$

$\dfrac{2\sin(30^\circ)\cos(10^\circ)}{\cos(10^\circ)}$

$\boxed{1}$

$\begin{aligned} 4 \sin 10^\circ + \color{#3D99F6}{\sqrt{3}} \tan 10^\circ & = 4 \sin 10^\circ + \color{#3D99F6}{2 \cos 30^\circ} \frac{\sin 10^\circ}{\cos 10^\circ} \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Since } \cos 30^\circ = \frac{\sqrt{3}}{2}} \\ & = 4 \sin 10^\circ + 2 (\color{#3D99F6}{4\cos^3 10^\circ - 3\cos 10^\circ}) \frac{\sin 10^\circ}{\cos 10^\circ} \quad \quad \space \space \small \color{#3D99F6}{\text{As } \cos (3\theta) =4\cos^3 \theta - 3\cos \theta} \\ & = 4 \sin 10^\circ + 2 \sin 10^\circ (4\cos^2 10^\circ - 3) \\ & = 4 \sin 10^\circ + 2 \sin 10^\circ (4 - 4\sin^2 10^\circ - 3) \\ & = 4 \sin 10^\circ + 2 \sin 10^\circ (1 - 4\sin^2 10^\circ ) \\ & = 6 \sin 10^\circ - 8\sin^3 10^\circ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{As } \sin (3\theta) =3\sin \theta - 4\sin^3 \theta} \\ & = 2\sin 30^\circ = \boxed{1} \end{aligned}$