## Precalculus (6th Edition) Blitzer

The exact value of the provided expression is $\sqrt{2}$.
We know that the reference angle for $\theta =\frac{7\pi }{4}$ is: \begin{align} & \theta '=2\pi -\frac{7\pi }{4} \\ & =\frac{8\pi -7\pi }{4} \\ & =\frac{\pi }{4} \end{align} Now, the function value for the reference angle is: $\sec \frac{\pi }{4}=\sqrt{2}$ Thus, we see that the angle $\frac{7\pi }{4}$ lies in quadrant IV, so the sec function is positive: \begin{align} & \sec \frac{7\pi }{4}=\sec \frac{\pi }{4} \\ & =\sqrt{2} \end{align}