Answer
The simplified form of the expression $\frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}$ is$\frac{2t+1}{{{t}^{2}}-1}$.
Work Step by Step
$\frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}$
Obtain the alternative form of the expression by multiplying the second term of rational expression $\frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}$ by $\frac{1+t}{1+t}$ and 1 in the form $\frac{-1}{-1}$:
$\begin{align}
& \frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}=\frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}.\frac{1+t}{1+t}.\frac{-1}{-1} \\
& =\frac{t}{{{t}^{2}}-1}-\frac{1+t}{1-{{t}^{2}}}.\frac{-1}{-1}
\end{align}$
Apply the Distributive property:
$\begin{align}
& \frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}=\frac{t}{{{t}^{2}}-1}-\frac{-\left( 1+t \right)}{-\left( 1-{{t}^{2}} \right)} \\
& =\frac{t}{{{t}^{2}}-1}-\frac{-\left( 1+t \right)}{-\left( 1-{{t}^{2}} \right)} \\
& =\frac{t}{{{t}^{2}}-1}+\frac{1+t}{{{t}^{2}}-1}
\end{align}$
Now, the denominators are same. So, add the numerators and keep the common denominator:
$\begin{align}
& \frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}=\frac{t}{{{t}^{2}}-1}+\frac{1+t}{{{t}^{2}}-1} \\
& =\frac{t+1+t}{{{t}^{2}}-1}
\end{align}$
Rearrange the terms and simplify further as follows:
$\begin{align}
& \frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}=\frac{t+1+t}{{{t}^{2}}-1} \\
& =\frac{2t+1}{{{t}^{2}}-1}
\end{align}$
