<s>[INST] Need help showing $\cos(x + y) / \sin(x - y) = -2$ if $\tan(\pi/4 - x) = 3 \tan(\pi/4 - y)$. I've tried manipulating the given relations using basic trig identities (which is all I know) but I wasn't able to prove anything.
 [/INST] $\tan(\pi/4 - x) = 3 \tan(\pi/4 - y)$
$\iff \frac{\sin(\pi/4 - x)}{\cos(\pi/4-x)}=\frac{3 \sin(\pi/4 - y)}{\cos(\pi/4-y)}$
$\iff \frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}=\frac{3 \cos(y)- 3\sin(y)}{\cos(y)+\sin(y)}$
$\iff \cos(x)\cos(y)+\sin(y)\cos(x)-\sin(x)\cos(y)-\sin(x)\sin(y)= 3\cos(x) \cos(y)+3\sin(x)\cos(y)-3\sin(y)\cos(x)-3\sin(x)\sin(y)$
$\iff 2\cos(x) \cos(y) + 4 \sin(x) \cos(y) - 4 \sin(y) \cos(x) - 2 \sin(x) \sin(y) = 0$
$\iff 2 \cos(x + y) + 4 \sin(x -y) = 0$
$\iff \frac{\cos(x + y)}{\sin(x - y)} = -2$
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