Precalculus (6th Edition) Blitzer

The required value of $\sec \left( {{\sin }^{-1}}\frac{1}{2} \right)$ is $\frac{2}{\sqrt{3}}$.
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 $\frac{2}{\sqrt{3}}$.