Answer
$$ \mathbf{T}=\frac{1}{\sqrt{2}} \mathbf{i}+\frac{1}{\sqrt{2}} \mathbf{j} $$
Work Step by Step
Given $$\mathbf{r}= e^t\cos t\mathbf{i} +e^t\mathbf{j} $$
Since
$$ \mathbf{r'}= e^t(\cos t-\sin t) \mathbf{i} +e^t\mathbf{j} $$
and $$ \| \mathbf{r'} \|=\sqrt{e^{2t}(\cos t-\sin t)^2+e^{2t} } =e^t \sqrt{ 2-\sin2t}$$
Then unit tangent vector given by
\begin{align*}
\mathbf{T}(t)&=\frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\\
&=\frac{ e^t(\cos t-\sin t) \mathbf{i} +e^t\mathbf{j} }{e^t \sqrt{2-\sin2t}}\\
&=\frac{ (\cos t-\sin t) \mathbf{i} + \mathbf{j} }{ \sqrt{ 2-\sin2t}}
\end{align*}
At $t=0$
$$ \mathbf{T}=\frac{1}{\sqrt{2}} \mathbf{i}+\frac{1}{\sqrt{2}} \mathbf{j} $$
