Exercise in algebra - Express different terms 
We have $ a = \dfrac{1}{\sqrt{1-b^2}}, c = \sqrt{\dfrac{1+b}{1-b}}, 0 \leq b < 1 $
Express $b$ in terms of $a$, $b$ in terms of $c$, $c$ in terms of $a$ and $a$ in terms of  $c$.

So I want to do this in the quickest and algebraically "cleanest" way. Here's my own attempt:
Part 1:
\begin{align*}
a\sqrt{1-b^2} &= 1\\
\sqrt{1-b^2} &= \dfrac{1}{a}\\
b^2 &= 1- \dfrac{1}{a^2}\\
b &= \sqrt{1 - \dfrac{1}{a^2}}
\end{align*}
Part 2:
\begin{align*}
c^2 &= \dfrac{1+b}{1-b}\\
b+1 &= c^2(1-b)\\
b+ bc^2 &= c^2 -1\\
b(c^2+1) &= c^2 - 1\\
b &= \dfrac{c^2-1}{c^2+1}
\end{align*}
3 and 4. I can't seem to do these. We want the denominator completely rationalized, etc. and I can't do this, I tried but my algebra becomes to hairy.
Are my solutions at 1/2 optimal, and how would we do 3/4 the best way? 

Now your 1 and 2 look fine.
For 3 : $$c=\frac{\sqrt{1+b}}{\sqrt{1-b}}\cdot\frac{\sqrt{1+b}}{\sqrt{1+b}}=\frac{1+b}{\sqrt{1-b^2}}=a(1+b)=a\left(1+\sqrt{1-\frac{1}{a^2}}\right)=a+\sqrt{a^2-1}.$$
For 4 : $$(c-a)^2=a^2-1\Rightarrow c^2-2ac=-1\Rightarrow a=\frac{c^2+1}{2c}.$$
