Answer
$\boldsymbol{T}(\frac{\pi}{4})=\langle\frac{\sqrt2}{6},\frac{-\sqrt2}{6},\frac{2\sqrt2}{3}\rangle$.
Work Step by Step
$\boldsymbol{r}(t)=\langle \sin^2{t},\cos^2{t},\tan^2{t}\rangle$
$\boldsymbol{T}(t)=\frac{\boldsymbol{r'}(t)}{|\boldsymbol{r'}(t)|}=\frac{1}{\sqrt{(2\sin{t}\cos{t})^2+(-2\sin{t}\cos{t})^2+(2\tan{t}\sec^2{t})^2}}\langle2\sin{t}\cos{t},-2\sin{t}\cos{t},2\tan{t}\sec^2{t}\rangle \Rightarrow \boldsymbol{T}(\frac{\pi}{4})=\frac{1}{\sqrt{(2\sin{(\frac{\pi}{4})}\cos{(\frac{\pi}{4})})^2+(-2\sin{(\frac{\pi}{4})}\cos{(\frac{\pi}{4})})^2+(2\tan{(\frac{\pi}{4})}\sec^2{(\frac{\pi}{4})})^2}}\langle2\sin{(\frac{\pi}{4})}\cos{(\frac{\pi}{4})},-2\sin{(\frac{\pi}{4})}\cos{(\frac{\pi}{4})},2\tan{(\frac{\pi}{4})}\sec^2{(\frac{\pi}{4})}\rangle=\frac{1}{3\sqrt2}\langle1,-1,4\rangle=\langle\frac{\sqrt2}{6},\frac{-\sqrt2}{6},\frac{2\sqrt2}{3}\rangle$
