If $\sin (B) = − \frac 1 2 $ with $B$ in third quadrant, then find $\cot (B/2)$

Question

If $\sin (B) = − \frac 1 2 $ with $B$ in third quadrant, then find $\cot (B/2)$

I'm getting $-\sqrt{3}-2$

Answer 1

$\cos B = -\dfrac{\sqrt{3}}{2}$, so $\cos(\frac{B}{2})$ = $-\sqrt{\dfrac{1 + \cos B}{2}} = -\dfrac{\sqrt{2 - \sqrt{3}}}{2}$, and $\sin(\frac{B}{2}) = \sqrt{\dfrac{1 - \cos B}{2}} = \dfrac{\sqrt{2 + \sqrt{3}}}{2}$, from here you can find $\cot(\frac{B}{2})$. From this we see that: $\cot(\frac{B}{2}) = ...= \sqrt{3} - 2$

Answer 2

Using Weierstrass substitution ,

$$\sin B=\frac{2\tan\dfrac B2}{1+\tan^2\dfrac B2}$$

$$\implies\frac{2\tan\dfrac B2}{1+\tan^2\dfrac B2}=-\frac12$$

Rearrange and solve the Quadratic Equation

Now as $\displaystyle180^\circ<B<270^\circ,90^\circ<\frac B2<135^\circ\implies \tan\dfrac B2<0$