Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\tan \alpha + \frac{\cos \alpha }{1+\sin\alpha }$ laat zich vereenvoudigen tot	A. $\boldsymbol{\frac{1}{\sin\alpha} }$
	B. $\boldsymbol{\frac{1}{\cos\alpha} }$
	C. $\boldsymbol{\frac{1}{\cos^2\alpha} }$
	D. $\boldsymbol{\frac{1}{\sin \alpha\,\cdot\,\cos\alpha} }$
	E. $\boldsymbol{\sin\alpha\cdot\cos\,\alpha }$
	F. $\boldsymbol{\frac{1}{\sin \alpha\,+\,\cos \alpha} }$
	G. $\boldsymbol{\sin\alpha+\cos\alpha }$

[ 6-2651 - op net sinds .2.15-(E)-2.11.2023 ]

Translation in E N G L I S H

... can be symplified to ...

Oplossing - Solution

$\\ \tan \alpha +\frac{\cos \alpha }{1+\sin \alpha }= \frac{\sin \alpha }{\cos \alpha } +\frac{\cos \alpha }{1+\sin \alpha }= \frac{\sin \alpha (1+\sin \alpha)+\cos^2 \alpha }{\cos \alpha \cdot (1+\sin \alpha )}\\= \frac{\sin \alpha +\sin^2 \alpha + \cos^2 \alpha }{\cos \alpha \cdot (1+\sin \alpha )}=\frac{\sin \alpha +1}{{\cos \alpha \cdot (1+\sin \alpha )}}=\frac{1}{\cos \alpha }$

$\tan \alpha + \frac{\cos \alpha }{1+\sin\alpha }$ laat zich vereenvoudigen tot	A. \(\boldsymbol{\frac{1}{\sin\alpha} }\)
	B. \(\boldsymbol{\frac{1}{\cos\alpha} }\)
	C. \(\boldsymbol{\frac{1}{\cos^2\alpha} }\)
	D. \(\boldsymbol{\frac{1}{\sin \alpha\,\cdot\,\cos\alpha} }\)
	E. \(\boldsymbol{\sin\alpha\cdot\cos\,\alpha }\)
	F. \(\boldsymbol{\frac{1}{\sin \alpha\,+\,\cos \alpha} }\)
	G. \(\boldsymbol{\sin\alpha+\cos\alpha }\)