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
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