Wiskunde Multiple Choice Vraag

Voor alle hoeken die geen veelvoud zijn van 90° is $\frac{\tan\alpha-\cot\alpha}{\tan\alpha+\cot\alpha}$ gelijk aan	A. − 1
B. − cos² α
C. − cos 2α
D. − tan² α
E. cos² α

1^ste manier :
$\frac{\tan{\alpha}-\cot{\alpha}}{\tan{\alpha}+\cot{\alpha}}=\frac{\tan{\alpha}-\frac{1}{\tan{\alpha}}}{\tan{\alpha}+\frac{1}{\tan{\alpha}}}=\frac{\tan^2{\alpha}-1}{\tan^2{\alpha}+1}=\frac{\tan^2{\alpha}-1}{\frac{1}{\cos^2{\alpha}}}=\cos^2{\alpha}.\tan^2{\alpha}-\cos^2{\alpha}=\sin^2{\alpha}-\cos^2{\alpha}=-\cos^2{\alpha}$

2^de manier :
$\frac{\tan{\alpha}-\cot{\alpha}}{\tan{\alpha}+\cot{\alpha}}=\frac{\frac{\sin{\alpha}}{\cos{\alpha}}-\frac{\cos{\alpha}}{\sin{\alpha}}}{\frac{\sin{\alpha}}{\cos{\alpha}}+\frac{\cos{\alpha}}{\sin{\alpha}}}=\frac{\sin^2{\alpha\;-\cos^2{\alpha}}}{\sin^2{\alpha+\\cos^2{\alpha}}}=\frac{-(\cos^2{\alpha\;}-\sin^2{\alpha)}}{1}=-\cos^2{\alpha}$