Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110} \frac {\tan \alpha} {1+tan^2\,\alpha}$ is gelijk aan	A. sin α
B. sin α . cos α
C. tan α + cot α
D. cos α ———— 1 + sin α
E. − tan α

$\frac {\tan \alpha} {1+tan^2\,\alpha}$ is equal to (see above)	A.
B.
C.
D.
E.

1^ste manier :
$\frac{\tan\alpha}{1+\tan^2\alpha}=\frac{\tan\alpha}{\frac{1}{\cos^2\alpha}}=\tan\alpha.\cos^2\alpha=\frac{\sin\alpha}{\cos\alpha}.\cos^2\alpha=\sin\alpha.\cos\alpha$
2^de manier :
$\\\frac{\tan\alpha}{1+\tan^2\alpha} =\frac{\frac{\sin\alpha}{\cos\alpha}}{1+\left(\frac{\sin\alpha}{\cos\alpha}\right)^2} =\frac{\frac{\sin\alpha}{\cos\alpha}}{1+\frac{\sin^2\alpha}{\cos^2\alpha}}\\ =\frac{\frac{\sin\alpha}{\cos\alpha}}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}=\frac{\frac{\sin\alpha}{\cos\alpha}}{\frac{1}{\cos^2\alpha}} =\frac{\sin\alpha}{\cos\alpha}.\frac{\cos^2\alpha}{1}=\sin\alpha.\cos\alpha$