Justify that $\tan 1^\circ$ is an irrational number

Here is a more algebraic approach not given in Minus One-Twelfth's link. Let $c$ and $s$ denote $\cos 1^\circ$ and $\sin 1^\circ$, respectively. We have $$(c+is)^{180}=e^{\pi i}=-1,$$ so $$\Im(c+is)^{180}=0.$$ That is $$\sum_{k=1}^{90}(-1)^{k-1}\binom{180}{2k-1}s^{2k-1}c^{180-2k+1}=0.$$ Note that $s,c\ne 0$. If $t$ is $\tan 1^\circ$, we have $$\sum_{k=1}^{90}(-1)^{k-1}\binom{180}{2k-1}t^{2(k-1)}=\frac{\sum_{k=1}^{90}(-1)^{k-1}\binom{180}{2k-1}s^{2k-1}c^{180-2k+1}}{sc^{179}}=0.$$

If $t$ is a rational number, then $t=\frac{p}{q}$ for some positive integers $p,q$ such that $\gcd(p,q)=1$. By the rational root theorem, $p,q\mid180$, so we can write $t=\frac{n}{180}$ for some integer $n$. Because $$\frac{\pi}{180}<\tan\frac{\pi}{180}<\frac{\tan(\pi/4)}{45}=\frac{1}{45},$$ where the first inequality is due to the inequality $\tan x>x$ for all $x\in(0,\pi/2)$, and the second inequality is true by convexity of $\tan$ on $(0,\pi/2)$. However, this means $$\pi<n<4,$$ but this is a contradiction (no integers lie strictly between $\pi$ and $4$). So $t=\tan 1^\circ$ cannot be rational.

Let $t_k = \tan k^\circ$ for positive integer $k$.

Please note that for $k < 45$, $0 < t_k < 1\implies 1 - t_k^2 \ne 0$.

Consider what happens if $t_1$ is ratonal. By repeat application of double-angle formula for tangent, we have

$$\begin{align} t_1 \in \mathbb{Q} \implies & t_2 = \frac{2t_1}{1-t_1^2} \in \mathbb{Q} \implies t_4 = \frac{2t_2}{1-t_2^2}\in \mathbb{Q}\\ \implies &t_8 = \frac{2t_4}{1-t_4^2}\in \mathbb{Q} \implies t_{16} = \frac{2t_8}{1-t_8^2}\in \mathbb{Q}\\ \implies &t_{32} = \frac{2t_{16}}{1 - t_{16}^2}\in \mathbb{Q} \end{align}$$ Recall $\displaystyle\;\tan(\theta -\phi) = \frac{\tan\theta - \tan\phi}{1 + \tan\theta\tan\phi}\;$, this implies $\;t_{30} = \frac{t_{32} - t_2}{1 + t_{32}t_2} \in \mathbb{Q}$. Since $t_{30} = \frac{1}{\sqrt{3}}$ is irrational, this is impossible.

As a result, $\tan 1^\circ = t_1$ is irrational.

(Another way).Let $\sin(1)=x$. We have $\sin(15)=\dfrac{\sqrt{2-\sqrt3}}{2}$. $$\sin(5(3))=5\sin(3)-20\sin^3(3)+16\sin^5(3)\\\sin(3)=3x-4x^3$$ Hence $$5(3x-4x^3)-20(3x-4x^3)^3+16(3x-4x^3)^5=\dfrac{\sqrt{2-\sqrt3}}{2}$$ If $x$ were rational then $(3x-4x^3)^n$ were rational for all integer $n$ so we would have $LHS$ rational and $RHS$ irrational, absurde.The irrationality of $\tan(1)$ follows.