Answer
The values are $\frac{1}{2},\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{1},\frac{2\sqrt{3}}{3}\,\text{ and }\,2$.
Work Step by Step
We have to use the Pythagorian Theorem to find the value of the base (b) for the given angle $\theta $
The perpendicular is $a=1$ and the hypotenuse is $c=2$.
So, the base $b$ is given by:
$\begin{align}
& {{a}^{2}}+{{b}^{2}}={{c}^{2}} \\
& {{b}^{2}}={{c}^{2}}-{{a}^{2}} \\
& ={{2}^{2}}-{{1}^{2}}
\end{align}$
Further simplify the equation:
$\begin{align}
& a=4-1 \\
& =\sqrt{3}
\end{align}$
Therefore, the base (b) is $\sqrt{3}$
$\begin{align}
& \sin \theta =\frac{a}{c} \\
& =\frac{1}{2}
\end{align}$
And the value of cos is:
$\begin{align}
& \cos \theta =\frac{b}{c} \\
& =\frac{\sqrt{3}}{2}
\end{align}$
And the value of tan is:
$\begin{align}
& \tan \theta =\frac{a}{b} \\
& =\frac{\sqrt{3}}{3}
\end{align}$
And the value of cot is:
$\begin{align}
& \cot \theta =\frac{b}{a} \\
& =\frac{\sqrt{3}}{1}
\end{align}$
And the value of sec is:
$\begin{align}
& \sec \theta =\frac{c}{b} \\
& =\frac{2}{\sqrt{3}} \\
& =\frac{2\sqrt{3}}{3}
\end{align}$
And the value of cosec is:
$\operatorname{cosec}\theta =\frac{c}{a}=2$
Thus, the values are $\frac{1}{2},\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{1},\frac{2\sqrt{3}}{3}\,\text{ and }\,2$.
