Answer
$${\bf{T}}\left( {\frac{\pi }{6}} \right) = \sqrt {\frac{3}{7}} {\bf{i}} - \frac{2}{{\sqrt 7 }}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 2\sin t{\bf{i}} + 4\cos t{\bf{j}},{\text{ }}t = \frac{\pi }{6} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {2\sin t{\bf{i}} + 4\cos t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2\cos t{\bf{i}} - 4\sin t{\bf{j}} \cr
& {\text{Evaluate at }}t = \frac{\pi }{6} \cr
& {\bf{r}}'\left( {\frac{\pi }{6}} \right) = 2\cos \left( {\frac{\pi }{6}} \right){\bf{i}} - 4\sin \left( {\frac{\pi }{6}} \right){\bf{j}} \cr
& {\bf{r}}'\left( {\frac{\pi }{6}} \right) = \sqrt 3 {\bf{i}} - 2{\bf{j}} \cr
& {\text{The unit tangent vector }}{\bf{T}}\left( t \right){\text{ at }}t{\text{ is defined as}} \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( t \right)} \right\|}} \cr
& {\text{at }}t = \frac{\pi }{6} \cr
& {\bf{T}}\left( {\frac{\pi }{6}} \right) = \frac{{{\bf{r}}'\left( {\pi /6} \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( {\pi /6} \right)} \right\|}} \cr
& {\bf{T}}\left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 {\bf{i}} - 2{\bf{j}}}}{{\left\| {\sqrt 3 {\bf{i}} - 2{\bf{j}}} \right\|}} = \frac{{\sqrt 3 {\bf{i}} - 2{\bf{j}}}}{{\sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( { - 2} \right)}^2}} }} = \frac{{\sqrt 3 {\bf{i}} - 2{\bf{j}}}}{{\sqrt 7 }} \cr
& {\bf{T}}\left( {\frac{\pi }{6}} \right) = \sqrt {\frac{3}{7}} {\bf{i}} - \frac{2}{{\sqrt 7 }}{\bf{j}} \cr} $$
