Question

If $5{\cos }^2\theta -2\sin \theta -2=0(\frac{5\pi }{4}<\theta <\frac{7\pi }{4})$
then prove that $\tan \frac{\theta }{2}=-1$

Answer 1

Changing to sine and factorizing, we have $(sin\theta +1)(5sin\theta -3)=0$. Since $\theta$ lies in the given interval, it must be -ive
$\therefore sin\theta =-1or\theta =\frac{3\pi }{2}=\frac{6\pi }{4}$
which lies in the given interval.
$\therefore \frac{\theta }{2}=\frac{3\pi }{4}$ and hence $tan\frac{\theta }{2}=tan\frac{3\pi }{4}=-1$