Wiskunde Multiple Choice Vraag

$Als\quad\frac{\sin 2x}{\sin x}=-1\\\\dan\;is\quad\frac{\tan 2x}{\tan x}=\;?$	A. 1
B. 2
C. −2
D. −1
E. \(\frac87\)

1^ste manier :
$\inline\\\frac{\sin 2x}{\sin x}=-1\Rightarrow\frac{2\sin x.\cos x}{\sin x}=-1\Rightarrow 2\cos x=-1\Rightarrow\cos x=-\frac12\Rightarrow\cos^2x=\frac14\\\\\quad\frac{\tan 2x}{\tan x}=\frac{2\tan x}{\tan x.(1-\tan^2 x)}=\frac{2}{1-\tan^2x}=\frac{2}{1-\left(\frac{1}{\cos^2x-1}\right)}=\frac{2}{2-\frac{1}{\cos^2x}}=\frac{2}{2-4}=-1$
2^de manier :
Als je voor x 120° neemt blijkt
$\inline\\\frac{\sin 2x}{\sin x}=\frac{\sin240^\circ}{\sin120^\circ}=\frac{-\sin60^\circ}{\sin60\circ}=-1\\\\\frac{\tan 2x}{\tan x}=\frac{\tan240^\circ}{\tan120^\circ}=\frac{\tan60^\circ}{-\tan60^\circ}=-1$