Answer
\begin{align*}
\mathbf{T} &=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}\\
\text{Length }& =1\end{align*}
Work Step by Step
Since
$$\mathbf{r}=(t \sin t+\cos t) \mathbf{i}+(t \cos t-\sin t) \mathbf{j}$$ Then \begin{align*}
\mathbf{v}&=(\sin t+t \cos t-\sin t) \mathbf{i}+(\cos t-t \sin t-\cos t) \mathbf{j}
\\
&=(t \cos t) \mathbf{i}-(t \sin t) \mathbf{j} \\
|\mathbf{v}|
&=\sqrt{(t \cos t)^{2}+(-t \sin t)^{2}}=\sqrt{t^{2}}\\
&=|t|=t
\end{align*}
Hence
\begin{align*}
\mathbf{T}&=\frac{\mathbf{v}}{|\mathbf{v}|}\\
&=\left(\frac{t \cos t}{t}\right) \mathbf{i}-\left(\frac{t \sin t}{t}\right) \mathbf{j}\\
&=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}\\ \text{Length }&=\int_{\sqrt{2}}^{2} t d t\\
&=\left[\frac{t^{2}}{2}\right]_{\sqrt{2}}^{2}=1\end{align*}
