Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The simplified form of the expression $\frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}$ is$\frac{2t+1}{{{t}^{2}}-1}$.
$\frac{t}{{{t}^{2}}-1}-\frac{1}{1-t}$ 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}