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?
 A: Your method is correct, you just forgot one subtlety:
$$\sin^2\beta +\cos^2\beta = 1\implies \cos\beta = \color{red}{\pm}\sqrt{1-\sin^2\beta}$$
We are given that $\pi<\beta<\frac{3\pi}{2}\implies \color{red}{\cos\beta <0}$
Therefore you should instead get $\cos\beta = \color{red}-\frac{2\sqrt{2}}{3}$ 
Subbing this in we get: $$h(\beta) = \frac{-\frac{2\sqrt{2}}{3}}{-\frac{2\sqrt{2}}{3}-1} =\frac{2\sqrt{2}}{2\sqrt{2}+3} =6\sqrt{2}-8\quad\text{as required}$$
A: You answer is very close.

Solve for$\beta$

Simplify
$$ 
\sin \left( \beta - \pi \right) = - \sin \beta = \frac{1}{3}
$$
The solutions are
$$
 \beta = - \arcsin \frac{1}{3} + 2\pi k, \quad \beta = \pi + \arcsin \frac{1}{3} + 2 \pi k
$$
where $k$ is an integer.
The constraint given in the problem is satisfied when
$$
 \boxed{\beta = \pi +\sin ^{-1}\left(\frac{1}{3}\right)}
$$

Compute the cosine
$$
\cos \beta = 
- \sqrt{1-\sin^{2}\beta} =
- \sqrt{1 -\left(\frac{1}{3}\right)^{2}}=  
-\frac{2 \sqrt{2}}{3}
$$
The external minus sign is required because of the constraint on $\beta$.

Write solution.

Substitute
$$
\frac{\cos \beta}{\cos \beta - 1} =
-\frac{2 \sqrt{2}}{3 \left(-\frac{2 \sqrt{2}}{3}-1\right)} =
-\frac{2 \sqrt{2}}{-3-2 \sqrt{2}} =
\boxed{6 \sqrt{2}-8}
$$
A: 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
