$\lim\limits_{x\to\large\frac{\pi}{2}}\large\frac{\sin(\cos x)\cos x}{\sin x-cosec x}$$=\lim\limits_{x\to \Large\frac{\pi}{2}}\large\frac{\sin x(\cos x)\cos x}{\sin x-\large\frac{1}{\sin x}}$
$\Rightarrow \lim\limits_{x\to\large\frac{\pi}{2}}\large\frac{\sin(\cos x)\cos x\sin x}{-\cos^2x}$
$\Rightarrow -\lim\limits_{x\to\large\frac{\pi}{2}}\large\frac{\sin(\cos x)\sin x}{-\cos x}$
$\lim\limits_{x\to\large\frac{\pi}{2}}\large\frac{\sin(\cos x)}{\cos x}$$=1$
$\Rightarrow -1$
Hence (b) is the correct answer.