Answer
The required value of $\sec \left( {{\sin }^{-1}}\frac{1}{2} \right)$ is $\frac{2}{\sqrt{3}}$.
Work Step by Step
Let $\theta $ represent the angle on $\left( -\frac{\pi }{2},\frac{\pi }{2} \right)$ as follows.
As $\sin \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ and, $\sec \theta =\frac{\text{Hypotenuse}\left( \text{H} \right)}{\text{Base}\left( \text{B} \right)}$.
Where, $\sin \theta $ is $\frac{1}{2}$.
$\begin{align}
& \theta ={{\sin }^{-1}}\frac{1}{2} \\
& \sin \theta =\frac{1}{2}
\end{align}$
Since, $\sin \theta $ is positive in the first quadrant.
$\begin{align}
& \sin \theta =\frac{1}{2}\text{ } \\
& =\frac{1}{2}\text{ } \\
& =\frac{\text{perpendicular}(y)}{\text{hypotenuse}(x)}
\end{align}$
By using the Pythagorian Theorem,
$\begin{align}
& {{z}^{2}}+{{y}^{2}}={{x}^{2}} \\
& ={{x}^{2}}-{{y}^{2}} \\
& =4-1
\end{align}$
Thus,
$z=\sqrt{3}$
Now
$\begin{align}
& \sec \left( {{\sin }^{-1}}\frac{1}{2} \right)=\sec \theta \\
& =\frac{\text{hypotenuse}(x)}{\text{base}(z)} \\
& =\frac{2}{\sqrt{3}}
\end{align}$
Thus, the value of $\sec \left( {{\sin }^{-1}}\frac{1}{2} \right)$ is $\frac{2}{\sqrt{3}}$.
