Let ,
$y=\lim_{{x->0}}(1-\sin x\cos x)^{cosec2x}$
$\log_{e}y=\lim_{{x->0}}\log_{e}(1-\sin x\cos x)^{cosec2x}$
$\log_{e}y=\lim_{{x->0}}cosec2x\log_{e}(1-\sin x\cos x)$
$\log_{e}y=\lim_{{x->0}}\frac{\log_{e}(1-\sin x\cos x)}{sin2x}$
$\log_{e}y=\lim_{{x->0}}\frac{\log_{e}(1-\sin x\cos x)}{2sinxcosx}$
$\log_{e}y=\frac{1}{2}\lim_{{x->0}}\frac{\log_{e}(1-\sin x\cos x)}{sinxcosx}$
Let , $z=\sin x\cos x$ ,
now as , $x\rightarrow 0$ then $z\rightarrow 0$ .
So,$\log_{e}y=\frac{1}{2}\lim_{z\rightarrow 0}\frac{\log_{e}(1-z)}{z}$
Now, $\lim_{z\rightarrow 0}\frac{\log_{e}(1-z)}{z}=-1$ [ Its easy so not calculating]
So,$\log_{e}y=\frac{1}{2}(-1)=-\frac{1}{2}$
So, $\large y=e^{-\frac{1}{2}}$ ………………….………….. (D) is the answer ..