Ans: C
$\begin{array}{cl}
& \dfrac{\cos 180^\circ}{1+ \sin (90^\circ + \theta)} + \dfrac{\cos 360^\circ}{1+ \sin (270^\circ + \theta)} \\
= & \dfrac{-1}{1 + \cos \theta} + \dfrac{1}{1 – \cos \theta} \\
= & \dfrac{-(1-\cos\theta) + (1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)} \\
= & \dfrac{2\cos\theta}{1-\cos^2\theta} \\
= & \dfrac{2\cos\theta}{\sin^2\theta}
\end{array}$
$\begin{array}{cl}
& \dfrac{\cos 180^\circ}{1+ \sin (90^\circ + \theta)} + \dfrac{\cos 360^\circ}{1+ \sin (270^\circ + \theta)} \\
= & \dfrac{-1}{1 + \cos \theta} + \dfrac{1}{1 – \cos \theta} \\
= & \dfrac{-(1-\cos\theta) + (1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)} \\
= & \dfrac{2\cos\theta}{1-\cos^2\theta} \\
= & \dfrac{2\cos\theta}{\sin^2\theta}
\end{array}$
