Answer
$${\bf{T}}\left( t \right) = \left\langle {\cos t, - \sin t} \right\rangle {\text{ and }}{\bf{N}}\left( t \right) = \left\langle { - \sin t, - \cos t} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {t,\ln \cos t} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}\left| {{\bf{v}}\left( t \right)} \right| \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \left\langle {1, - \tan t} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {1 + {{\tan }^2}t} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\sec }^2}t} = \sec t \cr
& \cr
& {\text{Find the unit tangent vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} \cr
& {\bf{T}}\left( t \right) = \frac{{\left\langle {1, - \tan t} \right\rangle }}{{\sec t}} \cr
& {\bf{T}}\left( t \right) = \left\langle {\cos t, - \sin t} \right\rangle \cr
& \cr
& {\text{Find the principal unit normal vector }}{\bf{N}}\left( t \right) \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left\langle {\cos t, - \sin t} \right\rangle \cr
& {\bf{T}}'\left( t \right) = \left\langle { - \sin t, - \cos t} \right\rangle \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {{{\sin }^2}t + {{\cos }^2}t} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = 1 \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} = \frac{{\left\langle { - \sin t, - \cos t} \right\rangle }}{1} \cr
& {\bf{N}}\left( t \right) = \left\langle { - \sin t, - \cos t} \right\rangle \cr} $$
