Reducing $\frac{\left(\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1\right)\cdot c}{2c\sqrt{\frac{a + b}{a - b}}}$ to $\frac{1}{\sqrt{1 - b^2/a^2}} $ I want to rearrange 
$$ \frac{\left(\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1\right) \cdot c}{2c \cdot \sqrt{\frac{a + b}{a - b}}} $$
into $$ \frac{1}{\sqrt{1 - b^2/a^2}} $$
I would be grateful for some hints or full explanation.
I do know you can get rid of $c$ and the root above easily.
 A: HINTS: After cancelling the $c$ as you mention, we have
\begin{align}
\frac{\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1}{2 \sqrt{\frac{a + b}{a - b}}}
= \frac{\left(\sqrt{\frac{a + b}{a - b}}\right)^2}{2 \sqrt{\frac{a + b}{a - b}}} + \frac{1}{{2 \sqrt{\frac{a + b}{a - b}}}} 
\end{align}
Now you can easily cancel stuff in the first term. Do you see how?
If you get stuck along the way, check out the spoilers by hovering your mouse over them (if you're on a computer).
Step 1: Canceling as mentioned above. 

 $$=\frac{\sqrt{\frac{a + b}{a - b}}}{2} + \frac{1}{{2 \sqrt{\frac{a + b}{a - b}}}}  = \frac{\sqrt{a + b}}{2\sqrt{{a - b}}} + \frac{\sqrt{a - b}}{2\sqrt{a + b}}$$

Step 2: Make denominators equal.

 $$=\frac{a + b}{2\sqrt{{(a - b)(a + b)}}} + \frac{a - b}{2\sqrt{(a + b)(a - b)}}$$

Step 3: Now add and cancel.

 $$=\frac{a + b + (a - b)}{2\sqrt{{(a - b)(a + b)}}} = \frac{2a}{2\sqrt{{(a - b)(a + b)}}}$$

Step 4: Bring everything under the square root.

 $$= \frac{a}{\sqrt{{a^2 - b^2}}} = \sqrt{\frac{a^2}{a^2 - b^2}}$$

Step 5: Divide to get the desired form.

 $$ = \sqrt{\frac{1}{\frac{a^2}{a^2} - \frac{b^2}{a^2}}} = \frac{1}{\sqrt{1 - b^2 / a^2}}.$$

