Chapter 6, Problem 40RE

To determine

Value of given trigonometric expression

Answer to Problem 40RE

  $\sec \frac{13\pi }{6}=-\frac{2}{\sqrt{3}}$

Explanation of Solution

Given information:

  $\sec \frac{13\pi }{6}$

Formula used:

  $\sec \left(n\left(\pi \right)+\theta \right)=-\sec \left(\theta \right)$

Where n is an odd number.

  $\begin{array}{l}\sec \frac{13\pi }{6}\\ =\sec \left(\frac{12\pi +\pi }{6}\right)\\ =\sec \left(\pi +\frac{\pi }{6}\right)\\ \end{array}$

As,

  $\begin{array}{l}\sec \left(n\left(\pi \right)+\theta \right)=-\sec \left(\theta \right)\\ \sec \left(\pi +\frac{\pi }{6}\right)=-\sec \frac{\pi }{6}\end{array}$

  $\boxed{\sec \frac{13\pi }{6}=-\sec \frac{\pi }{6}=-\frac{2}{\sqrt{3}}}$