Explanation

(b)
Given that, $\sin{\theta = - \frac{4}{5}and\ \theta\ }$lies in the IIIrd quadrant
$\Rightarrow \cos{\theta - \sqrt{1 - \frac{16}{25}}} = - \frac{3}{5}$
Now, $\cos{\frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos\theta}{2}}} = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}} = \pm \sqrt{\frac{1}{5}}$
But we take $\cos{\frac{\theta}{2} = - \frac{1}{\sqrt{5}}.Since,\ if\ \theta\ lies\ in\ IIIrd\ quadrant,\ }\ then\ \frac{\theta}{2}\$will be in $IInd\$quadrant
Hence,$\ \cos{\frac{\theta}{2} = - \frac{1}{\sqrt{5}}}$