Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

sin²α + sin²α.tan²α laat zich vereenvoudigen tot	A. 1
B. tan²α
C. 2 sin²α + 1
D. sin²α (1 + sin² α)
E. sin⁶α

The expression sin²α + sin²α.tan²α can be simplified to	A. 1
B. tan²α
C. 2 sin²α + 1
D. sin²α (1 + sin² α)
E. sin⁶α

$https://latex.codecogs.com/png.image?\dpi{110}\\{\color{DarkGreen} 1^{ste}\;manier\;: }\\ \sin^2\alpha +\sin^2\alpha .\tan^2\alpha =\sin^2\alpha .(1+\tan\alpha )=\sin^2\alpha \cdot \frac{1}{\cos^2\alpha } = \left ( \frac{\sin\alpha }{\cos\alpha } \right )^2 = ...$
$https://latex.codecogs.com/png.image?\dpi{110}\\{\color{DarkGreen} 2^{de}\;manier\;: }\\ \sin^2\alpha +\sin^2\alpha .\tan^2\alpha =\sin^2\alpha + \sin^2\alpha .\frac{\sin^2\alpha }{\cos^2\alpha }\\{\color{Green} .} \qquad\quad\quad=\frac{\sin^2\alpha .\cos^2\alpha +\sin^4\alpha }{\cos^2\alpha }=\frac{\sin^2\alpha .(\cos^2\alpha +\sin^2\alpha )}{\cos^2\alpha }=\frac{\sin^2\alpha }{\cos^2\alpha }=...$
$https://latex.codecogs.com/png.image?\dpi{110}\\{\color{DarkGreen} 2^{de}\;manier\;: }\\ \sin^2\alpha +\sin^2\alpha .\tan^2\alpha =\sin^2\alpha + \sin^2\alpha .\frac{\sin^2\alpha }{\cos^2\alpha }\\{\color{Green} .} \qquad\quad\quad=\frac{\sin^2\alpha .\cos^2\alpha +\sin^4\alpha }{\cos^2\alpha }=\frac{\sin^2\alpha .(\cos^2\alpha +\sin^2\alpha )}{\cos^2\alpha }=\frac{\sin^2\alpha }{\cos^2\alpha }=...$