The question can be expressed diagrammatically as
where
[tex]y=\frac{\sqrt[]{7}}{3}[/tex]
To find x, we can consider the triangle;
Using the Pythagorean Theorem, we have
[tex]\begin{gathered} 1^2=(\frac{\sqrt[]{7}}{3})^2+x^2 \\ 1=\frac{7}{9}+x^2 \\ x^2=1-\frac{7}{9}=\frac{2}{9} \\ x=\frac{\sqrt[]{2}}{3} \end{gathered}[/tex]
Next, we bring out the Trigonometric ratios:
[tex]\begin{gathered} \sin \theta=\frac{\sqrt[]{7}}{3} \\ \cos \theta=\frac{\sqrt[]{2}}{3} \\ \tan \theta=\frac{\sqrt[]{7}}{3}\div\frac{\sqrt[]{2}}{3}=\sqrt[]{\frac{7}{2}} \end{gathered}[/tex]
Therefore,
[tex]\frac{3}{\sqrt[]{7}}=\frac{1}{\sin\theta}=co\sec \theta[/tex]
and
[tex]\sqrt[]{\frac{7}{2}}=\tan \theta[/tex]
