## Trigonometry (11th Edition) Clone

$\sec \theta = -\frac{4}{3}$
Given $\tan\theta = \frac{\sqrt 7}{3}$ and $\theta$ is in quadrant III. Using Identity, $\sec^{2}\theta = 1 + \tan^{2}\theta$ $\sec^{2}\theta = 1 + (\frac{\sqrt 7}{3})^{2}$ $= 1 + (\frac{ 7}{9})$ $= \frac{9+7}{9} = \frac{16}{9}$ $\sec^{2}\theta = \frac{16}{9}$ $\sec \theta = +\frac{4}{3}$ or $-\frac{4}{3}$ In quadrant III $\sec \theta$ is negative. Therefore, $\sec \theta =-\frac{4}{3}$