Solution:
Given:
[tex]cot(\theta)=2[/tex]
Using the trig ratio of cot;
[tex]\begin{gathered} cot\theta=\frac{1}{tan\theta} \\ tan\theta=\frac{opposite}{adjacent} \\ Hence, \\ cot\theta=\frac{adjacent}{opposite} \\ cot\theta=\frac{2}{1} \\ adjacent=2 \\ opposite=1 \end{gathered}[/tex]
The hypotenuse is gotten using the Pythagoras theorem;
[tex]\begin{gathered} h^2=2^2+1^2 \\ h^2=4+1 \\ h^2=5 \\ h=\sqrt{5} \end{gathered}[/tex]
Thus, the sketch of the right triangle is;
[tex]\begin{gathered} Thus, \\ opposite=1 \\ adjacent=2 \\ hypotenuse=\sqrt{5} \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} sin\theta=\frac{opposite}{hypotenuse} \\ sin\theta=\frac{1}{\sqrt{5}} \\ sin\theta=\frac{1}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5} \\ sin\theta=\frac{\sqrt{5}}{5} \end{gathered}[/tex][tex]\begin{gathered} cos\theta=\frac{adjacent}{hypotenuse} \\ cos\theta=\frac{2}{\sqrt{5}} \\ cos\theta=\frac{2\sqrt{5}}{5} \end{gathered}[/tex][tex]\begin{gathered} tan\theta=\frac{opposite}{adjacent} \\ tan\theta=\frac{1}{2} \end{gathered}[/tex][tex]\begin{gathered} csc\theta=\frac{1}{sin\theta} \\ csc\theta=\frac{1}{\frac{\sqrt{5}}{5}} \\ csc\theta=\sqrt{5} \end{gathered}[/tex][tex]\begin{gathered} sec\theta=\frac{1}{cos\theta} \\ sec\theta=\frac{1}{\frac{2\sqrt{5}}{5}} \\ sec\theta=\frac{\sqrt{5}}{2} \end{gathered}[/tex]
