Proving $\tan 3^{\circ}$ is irrational

I want to show $$\tan 3^{\circ}$$ is irrational I have looked at other similar questions but they all have different patterns to approach trignometric values like $$\tan3$$ so what I am thinking of is assuming $$tan 3$$ is rational that is to say $$\tan 3 = a/b$$ so this means $$\sin 3^{\circ} / \cos 3^{\circ} = a/b$$ so we have $$\sin 3^{\circ}b= \cos 3^{\circ} a$$ so from here I want to find a contradiction but I need help.

Assume $$\tan(3^{\circ})$$ is rational that is $$\exists a,b\in\mathbb{N}$$, $$b\neq0$$ and $$\gcd(a,b)=1$$ such that $$\displaystyle\tan(3^{\circ})=\frac{a}{b}$$. Therefore : $$\tan(6^{\circ})=\frac{2\tan(3^{\circ})}{1-\tan^{2}(3^{\circ})} =\displaystyle\frac{\displaystyle\frac{2a}{b}}{\displaystyle\frac{b^{2}-a^{2}}{b^{2}}} =\frac{2ab}{b^{2}-a}\in\mathbb{Q}$$ Repeat this until you observe that $$\tan(12^{\circ})\in\mathbb{Q}\implies\tan(24^{\circ})\in\mathbb{Q}$$. However notice that : $$\tan(30^{\circ})=\tan(24^{\circ}+6^{\circ})=\frac{\tan(24^{\circ})+\tan(6^{\circ})}{1-\tan(24^{\circ})\tan(6^{\circ})}$$ Since $$\tan(24^{\circ}) ,\tan(6^{\circ})\in\mathbb{Q}$$, then $$\tan(30^{\circ})\in\mathbb{Q}$$. However : $$\tan(30^{\circ})=\frac{\sqrt{3}}{3}\notin\mathbb{Q}$$ Contradiction is thus achieved. Hence, $$\tan(3^{\circ})\notin\mathbb{Q}$$
• I am so glad to hear $\text{:-)}$ Feb 8 '21 at 8:57