Given:
[tex]\begin{gathered} \cos \alpha=-\frac{\sqrt[]{2}}{2} \\ \tan \alpha<0 \end{gathered}[/tex]To find the six trigonometric functions:
Since,
[tex]\cos \alpha<0\text{ and }\tan \alpha<0[/tex]So, the angle lies in the second quadrant.
Let us draw the diagram.
Using the Pythagoras theorem,
[tex]\begin{gathered} AB^2=AC^2+BC^2 \\ 2^2=AC^2+(\sqrt[]{2})^2 \\ 4=AC^2+2 \\ AC^2=2 \\ AC=\sqrt[]{2}\ldots\ldots\ldots(1) \end{gathered}[/tex]Next, find the six trig functions.
[tex]\begin{gathered} \sin \alpha=\frac{AC}{AB} \\ =\frac{\sqrt[]{2}}{2} \\ co\sec \alpha=\frac{AB}{AC} \\ =\frac{2}{\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =\frac{2\sqrt[]{2}}{2} \\ =\sqrt[]{2} \end{gathered}[/tex]And,
[tex]\begin{gathered} \cos \alpha=-\frac{BC}{AB}(Since,II\text{ quadrant)} \\ =-\frac{\sqrt[]{2}}{2} \\ \sec \alpha=-\frac{AB}{BC}(Since,II\text{ quadrant)} \\ =-\frac{2}{\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =-\frac{2\sqrt[]{2}}{2} \\ =-\sqrt[]{2} \end{gathered}[/tex]Similarly,
[tex]\begin{gathered} \tan \alpha=-\frac{AC}{BC}(Since,II\text{ quadrant)} \\ =-\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =-1 \\ \cot \alpha=-\frac{BC}{AC}(Since,II\text{ quadrant)} \\ =-\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =-1 \end{gathered}[/tex]Therefore, the answers are,
[tex]\begin{gathered} \sin \alpha=\frac{\sqrt[]{2}}{2} \\ co\sec \alpha=\sqrt[]{2} \\ \cos \alpha=-\frac{\sqrt[]{2}}{2} \\ \sec \alpha=-\sqrt[]{2} \\ \tan \alpha=-1 \\ \cot \alpha=-1 \end{gathered}[/tex]