Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\tan\alpha +\tan\beta }{\cot\alpha +\cot\beta }$ is gelijk aan	A. 1
B. tanα . tanβ
C. cotα . cotβ
D. tan²α . tan²β
E. cot²α . cot²β

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\tan\alpha +\tan\beta }{\cot\alpha +\cot\beta }$ is equal to	A. 1
	B. (tanα)(tanβ)
	C. (cotα)(cotβ)
	D. (tan²α)(tan²β)
	E. (cot²α)(cot²β)

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