How to show $ \tan \alpha $ is transcendental?

Question

I need to show that $ \tan \alpha $ is transcendental for each non-zero algebraic number $ \alpha $. Can any one give me an idea ? Thanks.

Answer 1

Suppose $ \alpha $ is a non zero algebraic number. Then $ i\alpha $ is also algebraic and $ \lbrace i\alpha , -i\alpha \rbrace $ is a set of distinct algebraic numbers. By the Lindemann-Weierstrass theorem for any non-zero algebraic numbers $ \beta_{1},\beta_{2} $ we have $ \beta_1 e^{i\alpha} + \beta_2 e^{-i\alpha}\ne 0 $. Obtain a contradiction assume that $ \tan \alpha $ is algebraic. Then $ (\tan \alpha -1) $ and $ (\tan \alpha +1) $ are also algebraic.

Note that $ \tan \alpha =\dfrac{e^{i\alpha}-e^{-i\alpha}}{e^{i\alpha}+e^{-i\alpha}}. $

Then $ (\tan \alpha -1)e^{i\alpha}+(\tan \alpha +1)e^{-i\alpha}=0 $. But this is a contradiction. Therefore $ \tan \alpha $ is transcendental.