Answer
$$
T(0)=\langle 0, 1, 0 \rangle
$$
$$
\begin{aligned}
\mathrm{N}(0) &=\left\langle \frac{-1}{\sqrt{2}},0 ,\frac{-1}{\sqrt{2}} )\right\rangle
\end{aligned}
$$
$$
\begin{aligned}
\mathbf{B}(0)=\left\langle \frac{-1}{\sqrt{2}},0 ,\frac{1}{\sqrt{2}} )\right\rangle\\
\end{aligned}
$$
Work Step by Step
We have the curve
$$
\mathbf{r}(t)=\left\langle\cos t_{,} \sin t, \ln \cos t\right\rangle
$$
At $t=0$, we find that:
$$
\mathbf{r}(0)=\left\langle\cos t_{,} \sin (0), \ln \cos (0)\right\rangle=
\left(1, 0, 0 \right)
$$
We first compute the ingredients needed for the unit normal vector:
$$
r^{\prime}(t) =\left\langle -\sin t, \cos t, -\tan t \right\rangle
$$
$\Rightarrow$
$$
\begin{aligned}
\left|r^{\prime}(t)\right|&=\sqrt{(- \sin t)^{2}+( \cos t)^{2}+(-\tan t)^{2}}\\
&=\sqrt{1+(\tan t)^{2}}\\
&=\left|\sec(t)\right|
\end{aligned}
$$
Assume $-\frac{\pi}{2}\lt t \lt \frac{\pi}{2} $; then $\left|r^{\prime}(t)\right|= \sec(t) $
Then, the unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{\sec(t) }\left\langle -\sin t, \cos t, -\tan t \right\rangle\\
&=\frac{\left\langle -\sin t, \cos t, -\tan t \right\rangle}{\sec(t)}\\
&=\left\langle -\sin t \cos t, (\cos t)^{2}, -\sin t \right\rangle
\end{aligned}
$$
at the given point, corresponding to $t=0$, we find that:
$$
\begin{aligned}
T(0) &=\left\langle -\sin (0) \cos (0), (\cos (0))^{2}, -\sin (0) \right\rangle \\
&=\left\langle 0, 1, 0 \right\rangle \\
\end{aligned}
$$
Then, we can find:
$$
\begin{aligned}
\mathbf{T}^{\prime}(t)&=\langle-[(\sin t)(-\sin t)+(\cos t)(\cos t)], 2(\cos t)(-\sin t),-\cos t) \\
&=\left\langle\sin ^{2} t-\cos ^{2} t,-2 \sin t \cos t,-\cos t\right\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
\left|\mathrm{T}^{\prime}(t)\right|& =\sqrt{\left(\sin ^{2} t-\cos ^{2} t\right)^{2}+(-2 \sin t \cos t)^{2}+(-\cos t)^{2}}
\end{aligned}
$$
Then, the principal unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
\mathrm{N}(t)&=\frac{\mathbf{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|} \\
&=\frac{\left\langle\sin ^{2} t-\cos ^{2} t,-2 \sin t \cos t,-\cos t\right\rangle}{\sqrt{\left(\sin ^{2} t-\cos ^{2} t\right)^{2}+(-2 \sin t \cos t)^{2}+(-\cos t)^{2}}} \\
& \ \ \ \ \ \text{corresponds to} \ \ t=0,\\\
\mathrm{N}(0) &=\frac{\left\langle\sin ^{2} (0)-\cos ^{2} (0),-2 \sin (0) \cos (0),-\cos (0)\right\rangle}{\sqrt{\left(\sin ^{2} (0)-\cos ^{2} (0)\right)^{2}+(-2 \sin(0) \cos(0))^{2}+(-\cos (0))^{2}}}\\
&=\frac{\left\langle -1,0 ,-1)\right\rangle}{\sqrt{1+0+1}}\\
&=\frac{\left\langle -1,0 ,-1)\right\rangle}{\sqrt{2}}\\
&=\left\langle \frac{-1}{\sqrt{2}},0 ,\frac{-1}{\sqrt{2}} )\right\rangle.
\end{aligned}
$$
The binormal vector $\mathbf{B}(0)$ is given by:
$$
\begin{aligned}
\mathbf{B}(0)=\mathbf{T}(0) \times \mathbf{N}(0)&=\left[\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 0 \\
\frac{-1}{\sqrt{2}}, &0 & \frac{-1}{\sqrt{2}}\\
\end{array}\right]\\
&=\left\langle \frac{-1}{\sqrt{2}},0 ,\frac{1}{\sqrt{2}} )\right\rangle.\\
\end{aligned}
$$
