You're on the right track with using the conjugate to get $\frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})}.$ As has been said in the comments, this is equal to $\frac{2}{(\sqrt{1+t}+\sqrt{1-t})}.$ One way to think about that is $\frac{2t}{t} = (2 \cdot t) \div t = 2 \cdot (t \div t) = 2 \cdot 1 = 2,$ as in Austin Mohr's comment. I tend to think about it slightly differently: $\frac{2t}{t}=\left(2\cdot t\right)\cdot\frac{1}{t}=2\cdot\left(t\cdot\frac{1}{t}\right)=2\cdot1=2$ (sticking to just multiplication, without division, using the Associative Property of Multiplication, and then using the fact that $t$ and $\frac{1}{t}$ are multiplicative inverses so that their product is $1$). Or, for this particular problem, $\begin{align} \frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})} &=\left(2\cdot t\right)\cdot\left(\frac{1}{t}\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}}\right) \\ &=2\cdot\left(t\cdot\frac{1}{t}\right)\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}} \\ &=2\cdot 1\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}} \\ &=\frac{2}{\sqrt{1+t}+\sqrt{1-t}}. \end{align}$