Me again, dunno wot I am gonna do wif maths! lol

Hello,

Let's see the denominator :
$\sin(2b)=2\cos(b)\sin(b)$

Hence $\sin(2b)-\sin(b)=\sin(b)(2 \cos(b)-1)$

Do you see the thing ? Do you see which formula for cos(2b) you will have to use ?

$\cos(2b)=2 \cos^2(b)-1$

So $\cos(2b)+\cos(b)=2 \cos^2(b)-1+\cos(b)$

Now this is the tricky thing :

$=2 \cos^2(b)+2 \cos(b)-\cos(b)-1=2 \cos(b)(\cos(b)+1)-(\cos(b)+1)$ $=(\cos(b)+1)(2 \cos(b)-1)$

How did I know it ? Because I know where I want to go :D

--------> $\frac{\cos(2b)+\cos(b)}{\sin(2b)-\sin(b)}=\frac{(\cos(b)+1){\color{red} (2 \cos(b)-1)}}{\sin(b) {\color{red} (2 \cos(b)-1)}}$

$=\frac{\cos(b)+1}{\sin(b)}$