Wiskunde Multiple Choice Vraag

Als tan(α + β) = 2 en tan α = , dan is tan β gelijk aan	A. \(\frac32\)
B. \(\frac34\)
C. \(-\frac34\)
D. \(\frac43\)
E. \(\frac12\)

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°
$\inline\\\tan(90^\circ-\beta)=\tan 2\alpha\\\Leftrightarrow\cot\beta=\frac{2.\tan\alpha}{1-\tan^2\alpha}\\\Leftrightarrow\tan\beta=\frac{1-\left(\frac12\right)^2}{2.\frac12}\\\Leftrightarrow\tan\beta=1-\frac14=\frac34$

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