2025/004) prove that $\tan \, 1^{\circ}$ is irrational

 We know $\tan\,2A = \frac{2\tan\, A}{1+\tan ^2 A}\cdots(1)$

So if $\tan\, A $ is rational then $\tan 2A$ is rational 

Further $\tan(A+B) = \frac{\tan \, A + \tan\, B}{1-\tan\,A \tan \,B}\cdots(2)$

So if $\tan\, A$ and $\tan \, B$ are rational then $\tan(A+B)$ is rational

So if $\tan\, A$ and $\tan \,nA$ are rational then $\tan(n+1)A$ is rational

using Above 2 we get if $\tan\, A $ is rational then $\tan nA$ is rational for all $n \in Z$

if $\tan \, 1^{\circ}$ is rational then $\tan \, 30^{\circ}$ or $\frac{1}{\sqrt{3}}$ is rational which is not 

Hence $\tan \, 1^{\circ}$ is irrational 

Proved