<s>[INST] Find the exact value of $\frac{\cos( \beta)}{\cos( \beta) -1}$ with $\sin(\beta - \pi) = \frac{1}{3}$ 
Let $h$ be  $$\frac{\cos( \beta)}{\cos( \beta) -1}$$
With $\sin(\beta-\pi)=\frac{1}{3}$ and $\beta \in
 {]}\pi,\frac{3\pi}{2}{[}$ determine the exact value of $h(\beta)$.

I tried:
$$\sin(\beta-\pi) = \sin(\beta)\cos(\pi)-\cos(\beta)\sin(\pi) = -\sin(\beta)$$
$$\\$$
$$-\sin(\beta) = \frac{1}{3} \Leftrightarrow \\
\sin(\beta) = - \frac{1}{3}$$
$$\\$$
$$\cos \beta = \sqrt{1-\sin^2(\beta)} = \sqrt{1-(-\frac{1}{3})^2} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$$
And so $h(\beta)$ is :
$$\frac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} = \\
\frac{\sqrt{2}}{2\sqrt{2}-3}$$
But my book says the solution is $6\sqrt{2}-8$. What went wrong?
 [/INST] 3 things
$\dfrac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} = 
\dfrac{2\sqrt{2}}{2\sqrt{2}-3}$
you dropped a 2.
$\dfrac{2\sqrt{2}}{2\sqrt{2}-3} = -6\sqrt 2  - 8$
With a little bit of simplification you would see that you were "only" off by a minus sign.
and here is where you lost it.
$\cos \beta = -\frac {\sqrt {2}}{3}$ as $\beta$ is in QIII
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