The derivative of the vector is,

\(r'(t)=\{-\sin 5, \cos t, -\tan t \}\)

The unit tangent vector, T at the given point is obtained as follows.

\(T(t)=\frac{r'(t)}{||r'(t)||}\)

\(T(t)=\frac{\langle-\sin t,\cos t,-\tan t\rangle}{\sqrt{(-\sin t)^2+(\cos t)^2+(-\tan t }^2}\)

\(T(t)=\langle-\frac{\sin t}{\sec t},\frac{cos t}{sec t},-\frac{tan t}{\sec t}\rangle\)

\(T(t)|_{1,0,0}=\langle-0.45,1,0\rangle\)

The derivative of the unit tangent vector, T is,

\(\displaystyle{T}'{\left({t}\right)}={\left\langle-\frac{{ \cos{{t}}- \tan{{t}} \sin{{t}}}}{ \sec{{t}}},- \sin{{2}}{t},- \cos{{t}}\right\rangle}\)

The normal unit vector, N at the given point is obtained as follows.

\(\displaystyle{N}{\left({t}\right)}=\frac{{{T}'{\left({t}\right)}}}{{{\left|{\left|{T}'{\left({t}\right)}\right|}\right|}}}\)

\(\displaystyle{N}{\left({t}\right)}=\frac{{{\left\langle-\frac{{ \cos{{t}}- \tan{{t}} \sin{{t}}}}{ \sec{{t}}},- \sin{{2}}{t},- \cos{{t}}\right\rangle}}}{{\frac{{\sqrt{{{\left(- \cos{{t}}+ \tan{{t}} \sin{{t}}\right)}^{2}+{{\sin}^{2}{\left({2}{t}\right)}}{{\sec}^{2}{\left({t}\right)}}+{1}}}}}{ \sec{{\left({t}\right)}}}}}\)

\(\displaystyle{N}{\left({t}\right)}{|}_{{{1},{0},{0}}}={\left\langle{0.366},{0},-{0.70}\right\rangle}\)

The binormal vector, B at the given point is obtained as follows.

\(B(t)=T(t) xx N(t)\)

\(\displaystyle{B}{\left({t}\right)}{|}_{{{1},{0},{0}}}={\left[\begin{matrix}{i}&{j}&{k}\\-{0.45}&{1}&{0}\\{0.366}&{0}&-{0.70}\end{matrix}\right]}\)

\(=(-0.70)i-(0.315)j-(0.366)k\)