Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\cos 2x}{1-\tan^2x}$ is gelijk aan ( Uiteraard voor die hoeken waarvoor de noemer niet nul is)	A. sin²x
B. cos²x
C. tan²x
D. cot²x
E. sec²x

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\cos 2x}{1-\tan^2x}$ is equal to	A. sin²x
	B. cos²x
	C. tan²x
	D. cot²x
	E. sec²x

1^ste manier :
$\frac{\cos2\alpha}{1-\tan^2\alpha}=\frac{\cos^2\alpha-\sin^2\alpha}{1-\frac{\sin^2\alpha}{\cos^2\alpha}}=\frac{\cos^2\alpha-\sin^2\alpha}{\frac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}=\cos^2\alpha$
2^de manier : rekening houdend met
$1+\tan^2\alpha=\frac {1} {\cos^2\alpha}\;\;en\;\;1+\cos2\alpha=2\cos^2\alpha$
$\frac{\cos2\alpha}{1-\tan^2\alpha}=\frac{\cos2\alpha}{1-\left(\frac{1}{\cos^2\alpha-1}\right)}=\frac{\cos2\alpha}{2-\frac{1}{\cos^2\alpha}}=\frac{\cos2\alpha}{\frac{2\cos^2\alpha-1}{\cos^2\alpha}}=\frac{\cos2\alpha}{\frac{\cos2\alpha}{\cos^2\alpha}}=\cos^2\alpha$