This type of question might be voted down or frowned upon, but if anyone could point me in the right direction I would be very grateful. I have worked through a question getting the following result:
$$ \frac{\nu{'}}{\nu}=\frac{(\hat{\Omega}\cdot{\beta}-1)}{(\hat{\Omega{'}}\cdot{\beta}-1)+\frac{h\nu}{\gamma{m_e}c^2}(\hat{\Omega}\cdot\hat{\Omega}'-1)} $$
Now, the problem is: I think I have done this question correctly. That is my assumption. However, the question asks to show that:
$$ \frac{\nu{'}}{\nu}=\frac{(1-\hat{\Omega}\cdot{\beta})}{(1-\hat{\Omega{'}}\cdot{\beta})+\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')} $$
So my question is this: have I gone wrong or am I able to re-arrange my expression into the expression I need to show...?
To get
$$\frac{\nu{'}}{\nu}=\frac{(\hat{\Omega}\cdot{\beta}-1)}{(\hat{\Omega{'}}\cdot{\beta}-1)+\frac{h\nu}{\gamma{m_e}c^2}(\hat{\Omega}\cdot\hat{\Omega}'-1)}$$
changed into the correct form, we have $(\hat{\Omega}\cdot{\beta}-1)=-(1-\hat{\Omega}\cdot{\beta})$ and $(\hat{\Omega{'}}\cdot{\beta}-1)+\frac{h\nu}{\gamma{m_e}c^2}(\hat{\Omega}\cdot\hat{\Omega}'-1)=-(1-\hat{\Omega{'}}\cdot{\beta})-\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')$, and so we have
$$\frac{\nu{'}}{\nu}=\frac{-(1-\hat{\Omega}\cdot{\beta})}{-(1-\hat{\Omega{'}}\cdot{\beta})-\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')}=\frac{(1-\hat{\Omega}\cdot{\beta})}{(1-\hat{\Omega{'}}\cdot{\beta})+\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')}$$
where we have used the distributive property to extract $-1$ in the denominator.