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 |
[ 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 :
\(\large\frac{\cot\alpha\,+\,\cot\beta}{\tan\alpha\,+\,\tan\beta}=\frac{\frac{\cos\alpha}{\sin\alpha}+\frac{\cos\beta}{\sin\beta}}{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}=\frac{\frac{\cos\alpha.\sin\beta+\cos\beta.\sin\alpha}{\sin\alpha.\sin\beta}}{\frac{\sin\alpha.\cos\beta+\sin\beta.\cos\alpha}{\cos\alpha.\cos\beta}}\\=\frac{\cos\alpha.\sin\beta+\cos\beta.\sin\alpha}{\sin\alpha.\sin\beta}.\frac{\cos\alpha.\cos\beta}{\sin\alpha.\cos\beta+\sin\beta.\cos\alpha}\\=\frac{\cos\alpha.\cos\beta}{\sin\alpha.\sin\beta}=\frac{\cos\alpha}{\sin\alpha}.\frac{\cos\beta}{\sin\beta}=\cot\alpha.\cot\beta\)