Trigonometric Functions

We substitute $\tan \theta = -\frac{4}{3}$ into the identity: $\sec^2\theta = \tan^2\theta + 1 = \left(-\frac{4}{3}\right)^2 + 1 = \frac{25}{9}$ . Since $\sec \theta = \frac{1}{\cos \theta}$ , thus $\cos^2 \theta = \frac{9}{25}$ . Since $\frac{\pi}{2} < \theta < \pi$ , we take the negative root and $\cos \theta = -\frac{3}{5}$ .

By definition, $\sin \theta = \cos \theta \tan \theta = \left(-\frac{3}{5}\right) \cdot \left(-\frac{4}{3}\right) = \frac{4}{5}$ .

Therefore $\frac{1}{\sin \theta + \cos \theta} = \frac{1}{\frac{4}{5} -\frac{3}{5}} = 5$ .