Question

Compute $\tan \frac{\theta }{4}$ if $\cos \theta =-0.8$ and ${180}^{\circ }<\theta <{270}^{\circ }$

Answer 1

from the given conditions .we see that $45$${}^{\circ }$ <$\frac{\theta }{4}$ < ${67}^{\circ }30$ so that $\tan$ $\left(\frac{\theta }{4}\right)>0$
and ${90}^{\circ }<\frac{\theta }{2}<{135}^{\circ }\therefore \tan \left(\frac{\theta }{2}\right)=-ive$
$\tan \frac{\theta }{2}=\sqrt{\left(\frac{1-\cos \theta }{1+\cos \theta }\right)}=-\sqrt{\left(\frac{1+\frac{4}{5}}{1-\frac{4}{5}}\right)}=-3$
$\therefore$ Now $\tan$ $\frac{\theta }{2}=\frac{2t}{1-t^2}=-3wheret=\tan \frac{\theta }{4}$
$\therefore 3t^2-2t-3=0\therefore t=\frac{2\pm \sqrt{(4+36)}}{6}$
$\therefore t=\tan \frac{\theta }{4}=\frac{1+\sqrt{10}}{3}as\tan \frac{\theta }{4}$ is + ive