is gelijk aan

(uiteraard voor alle  α, β
die geen veelvoud zijn van 90°)
A.     1
B.   tan α.tan β
C.   cot α.cot β
D.   tan2α + tan2β
E.   cot2α + cot2β
A    B    C    D    E

[ 3-6854 - op net sinds 27.12.14-(E)-25.9.2024 ]

Translation in   E N G L I S H



is equal to
A.     1
B.   tanα.tanβ
C.   tanα.tanβ
D.   tan2α + tan2β
E.   cot2α + cot2β

Oplossing - Solution

1ste manier :
\(\frac{\cot\alpha\,+\,\cot\beta}{\tan\alpha\,+\,\tan\beta}=\frac{\frac{1}{\tan\alpha}+\frac{1}{\tan\beta}}{\tan\alpha\,+\,\tan\beta}=\frac{\frac{\tan\beta\,+\,\tan\alpha}{\tan\alpha\,.\,\tan\beta}}{\tan\alpha\,+\,\tan\beta}\\=\frac{\tan\beta\,+\,\tan\alpha}{\tan\alpha\,.\,\tan\beta}.\frac{1}{\tan\alpha\,+\,\tan\beta}=\frac{1}{\tan\alpha\,.\,\tan\beta}=\cot\alpha.\cot\beta\)
2de manier :
\(\large\frac{\cot\alpha\,+\,\cot\beta}{\tan\alpha\,+\,\tan\beta}=\frac{\cot\alpha\,+\,\cot\beta}{\frac{1}{\cot\alpha}+\frac{1}{\cot\beta}}=\frac{\cot\alpha\,+\,\cot\beta}{\frac{\cot\beta\,+\,\cot\alpha}{\cot\alpha\,.\,\cot\beta}}\\=(\cot\alpha\,+\,\cot\beta).\frac{\cot\alpha\,.\,\cot\beta}{\cot\beta\,+\,\cot\alpha}=\cot\alpha\,.\,\cot\beta\)
3de manier :
\(\large\frac{\cot\alpha\,+\,\cot\beta}{\tan\alpha\,+\,\tan\beta}=\frac{\frac{\cos\alpha}{\sin\alpha}+\frac{\cos\beta}{\sin\beta}}{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}=\frac{\frac{\cos\alpha.\sin\beta+\cos\beta.\sin\alpha}{\sin\alpha.\sin\beta}}{\frac{\sin\alpha.\cos\beta+\sin\beta.\cos\alpha}{\cos\alpha.\cos\beta}}\\=\frac{\cos\alpha.\sin\beta+\cos\beta.\sin\alpha}{\sin\alpha.\sin\beta}.\frac{\cos\alpha.\cos\beta}{\sin\alpha.\cos\beta+\sin\beta.\cos\alpha}\\=\frac{\cos\alpha.\cos\beta}{\sin\alpha.\sin\beta}=\frac{\cos\alpha}{\sin\alpha}.\frac{\cos\beta}{\sin\beta}=\cot\alpha.\cot\beta\)