Find the Exact Value sec(arctan(- square root of 3))

Draw a triangle in the plane with vertices ${\displaystyle (1,-\sqrt{3})}$, ${\displaystyle (1,0)}$, and the origin. Then ${\displaystyle \arctan (-\sqrt{3})}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (1,-\sqrt{3})}$. Therefore, ${\displaystyle \sec \left(\arctan (-\sqrt{3})\right)}$ is ${\displaystyle \sqrt{1+{(-\sqrt{3})}^2}}$.

Apply the product rule to ${\displaystyle -\sqrt{3}}$.

Raise ${\displaystyle -1}$ to the power of ${\displaystyle 2}$.

Multiply ${\displaystyle {\sqrt{3}}^2}$ by ${\displaystyle 1}$.

Use ${\displaystyle \sqrt[n]{a^x}=a^{\frac{x}{n}}}$ to rewrite ${\displaystyle \sqrt{3}}$ as ${\displaystyle 3^{\frac{1}{2}}}$.

Apply the power rule and multiply exponents, ${\displaystyle {\left(a^m\right)}^n=a^{mn}}$.

Combine ${\displaystyle \frac{1}{2}}$ and ${\displaystyle 2}$.

Cancel the common factor of ${\displaystyle 2}$.

Evaluate the exponent.

Add ${\displaystyle 1}$ and ${\displaystyle 3}$.

Rewrite ${\displaystyle 4}$ as ${\displaystyle 2^2}$.