Answer
$$ \mathbf{T}=\frac{-\sqrt{3}}{2 } \mathbf{i} +\frac{1}{2} \mathbf{j} $$
Work Step by Step
Given $$\mathbf{r}= 5\cos t\mathbf{i} +5\sin t\mathbf{j} $$
Since
$$ \mathbf{r'}= -5\sin t\mathbf{i} +5\cos t\mathbf{j}$$
and $$ \| \mathbf{r'} \|=\sqrt{ 25\sin^2 t+25\cos^2 t} =5$$
Then unit tangent vector given by
\begin{align*}
\mathbf{T}(t)&=\frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\\
&=\frac{ -5\sin t\mathbf{i} +5\cos t\mathbf{j}}{5}\\
&=- \sin t\mathbf{i} + \cos t\mathbf{j}
\end{align*}
At $t=\pi/3$
$$ \mathbf{T}=\frac{-\sqrt{3}}{2 } \mathbf{i} +\frac{1}{2} \mathbf{j} $$
