Obtuse calculation

From the identity $\sec^2 \theta = 1 + \tan^2 \theta$ , we have $\cos^2 \theta = \frac{1}{1 + \tan^2\theta} = \frac{1}{1 + \left(\frac{4}{3}\right)^2} = \frac{9}{25}$ . Since $\pi < \theta < \frac{3}{2}\pi$ , from the CAST rule, we have $\cos \theta = -\frac{3}{5}$ . From the double angle formula, we have $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1 = 1 - 2\sin^2 \frac{\theta}{2}$ , thus

$\begin{aligned} \sin^2\frac{\theta}{2} &= \frac{1-\cos \theta}{2} = \frac{1 - \left(-\frac{3}{5}\right)}{2} = \frac{4}{5} \\ \cos^2\frac{\theta}{2} &= \frac{1+\cos \theta}{2} = \frac{1 + \left(-\frac{3}{5}\right)}{2} = \frac{1}{5} \\ \tan^2\frac{\theta}{2} &= \frac{1-\cos \theta}{1+\cos \theta} = \frac{1 - \left(-\frac{3}{5}\right)}{1 + \left(-\frac{3}{5}\right)} = 4 \\ \end{aligned}$

Since $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3}{4}\pi$ , by the CAST rule, we have $\sin \frac{\theta}{2} > 0$ , $\cos \frac{\theta}{2} < 0$ and $\tan \frac{\theta}{2} < 0$ . Thus $\sin \frac{\theta}{2} = \frac{2}{\sqrt{5}}$ , $\cos \frac{\theta}{2} = - \frac{1}{\sqrt{5}}$ and $\tan \frac{\theta}{2} = - 2$ .

Hence $\sin \frac{\theta}{2} + 2 \cos \frac{\theta}{2} - 5\tan \frac{\theta}{2} = \frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}} -5(-2) = 10$ .