(b) Since ${\rm{cose}}{{\rm{c}}^2}\theta = 1 + {\cot ^2}\theta = 1 + \frac{9}{{16}} = \frac{{25}}{{16}}$

$\left( \because   {\tan \theta  =  – \frac{4}{3}} \right)$

${\sin ^2}\theta = \frac{1}{{{\rm{cose}}{{\rm{c}}^2}\theta }} = \frac{{16}}{{25}} $

$\Rightarrow \sin \theta = \pm \frac{4}{5},$

Both the values are acceptable, since $\tan \theta = – \frac{4}{3}\,\,$

$\,i.e.,\theta $ lies in ${2^{nd}}$ or ${4^{th}}$ quadrant.