1^ste manier :
\(2=\tan(\alpha+\beta)=\frac{\tan \alpha+\tan\beta}{1-\tan\alpha-\tan\beta} =\frac{\frac12+\:\tan\beta}{1-\frac12\tan\beta}=\frac{1\,+\,2\tan\beta}{2\,-\,\tan\beta}\)
Hieruit volgt :  4 − 2.tan β = 1 + 2.tan β   zodat
3 = 4.tan β  ⇔  tan β = \(\frac43\)
2^de manier :
\(\tan(\alpha+\beta)=2=\frac{1}{\frac12}=\frac{1}{\tan\alpha}=\cot\alpha=\tan(90^\circ-\alpha)\)
 ⇔ α + β = 90° − α + k.180°  ⇔  90° − β = 2.α − k.180°
\(\tan(90^\circ-\beta)=\tan 2\alpha\\\Leftrightarrow\cot\beta=\frac{2.\tan\alpha}{1-\tan^2\alpha}\\\Leftrightarrow\tan\beta=\frac{1-\left(\frac12\right)^2}{2.\frac12}\\\Leftrightarrow\tan\beta=1-\frac14=\frac34\)

3^de manier :
\(\tan\beta=\tan\left[(\alpha+\beta)-\alpha\right]=\frac{\tan(\alpha+\beta)-\tan\alpha}{1+\tan(\alpha+\beta).\tan\alpha}=\frac{2-\frac12}{1+1}=\frac{\frac32}{2}=\frac34\)