We have $\sqrt{\large\frac{1+\cos\theta}{1-\cos^2\theta}}$
$\Rightarrow \sqrt{\large\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}$
$\Rightarrow \large\frac{1+\cos\theta}{\sqrt{\sin^2\theta}}$
$\large\frac{1+\cos\theta}{\mid \sin\theta\mid}=\left\{\begin{array}{1 1}\large\frac{1+\cos\theta}{\sin\theta}&\normalsize if\;0\;<\;\theta\;<\pi\\\large\frac{1+\cos\theta}{-\sin\theta}&\normalsize if\;\pi\;<\;\theta\;<2\pi\end{array}\right.$
$\Rightarrow \left\{\begin{array}{1 1}cosec\theta+\cot\theta&if\;0<\theta>\pi\\-cosec\theta-\cot\theta&if\;\pi<\theta<2\pi\end{array}\right.$
Hence (c) is the correct answer.