Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

sin α + cos α sin α − cos α is gelijk aan	A. − 1
B. tan (α − 45°)
C. tan (α + 45°)
D. tan (45° − α)
E. − tan (α + 45°)

sin α + cos α sin α − cos α is equal to	A. − 1
B. tan (α − 45°)
C. tan (α + 45°)
D. tan (45° − α)
E. − tan (α + 45°)

1^ste manier :
$\\\frac{\sin{\alpha}+\cos{\alpha}}{\sin{\alpha}-\cos{\alpha}}=\frac{\sin+\sin(90^\circ-\alpha)}{\sin\alpha-\sin(90^\circ-\alpha)}=\frac{2\sin\frac{\alpha+90^\circ-\alpha}{2}.\cos\frac{\alpha-90^\circ+\alpha}{2}}{2\cos\frac{\alpha+90^\circ-\alpha}{2}.\sin\frac{\alpha-90^\circ+\alpha}{2}}\\=\frac{\sin45^\circ.\cos(\alpha-45^\circ)}{\cos45^\circ.\sin(\alpha-45^\circ)}=\frac{\cos(\alpha-45^\circ)}{\sin(\alpha-45^\circ)}=\cot(\alpha-45^\circ)=\\\tan(90^\circ-\alpha+45^\circ)=\tan(135^\circ-\alpha)=\tan(-45^\circ-\alpha)=-\tan(\alpha+45^\circ)$
2^de manier :
$\\\frac{\sin{\alpha}+\cos{\alpha}}{sin{\alpha}-cos{\alpha}}\!=\frac{tan{\alpha}\!+\!1}{tan{\alpha}\!-\!1}=-\frac{\tan{\alpha}\!+\!1}{1\!-\!\tan{\alpha}}=-\frac{\tan+\tan45^\circ}{1\!-\!\tan\alpha.\tan45^\circ}=-\tan(\alpha\!+\!45^\circ)$