If $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$, then the value of $\sin x$ is

Given the expression $x = \tan^{-1}\left(\frac{1}{\tan(\frac{\theta}{2})}\right) - \tan^{-1}\left(\tan(\frac{\theta}{2})\right)$.

Using the identity $\tan^{-1}(y) + \tan^{-1}\left(\frac{1}{y}\right) = \frac{\pi}{2}$ for $y > 0$
where $y = \tan(\frac{\theta}{2})$, we have:
$\tan^{-1}\left(\frac{1}{\tan(\frac{\theta}{2})}\right) = \frac{\pi}{2} - \tan^{-1}\left(\tan(\frac{\theta}{2})\right)$

Substituting into the expression for $x$:
$x = \left(\frac{\pi}{2} - \tan^{-1}\left(\tan(\frac{\theta}{2})\right)\right) - \tan^{-1}\left(\tan(\frac{\theta}{2})\right)$
$x = \frac{\pi}{2} - 2\tan^{-1}\left(\tan(\frac{\theta}{2})\right)$

Considering the given alternative form $\cot^{-1}(\sqrt{\cos\alpha}) - \tan^{-1}(\sqrt{\cos\alpha}) = x$
Using $\cot^{-1} y = \frac{\pi}{2} - \tan^{-1} y$:
$x = \left(\frac{\pi}{2} - \tan^{-1}(\sqrt{\cos\alpha})\right) - \tan^{-1}(\sqrt{\cos\alpha})$
$x = \frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos\alpha})$

To find $\sin x$:
$\sin x = \sin\left(\frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos\alpha})\right)$
Using $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$:
$\sin x = \cos\left(2\tan^{-1}(\sqrt{\cos\alpha})\right)$

Let $y = \sqrt{\cos\alpha}$ and $\theta = \tan^{-1}(y)$, so $\tan\theta = y$
Using the identity $\cos(2\theta) = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$:
$\sin x = \frac{1 - y^2}{1 + y^2} = \frac{1 - \cos\alpha}{1 + \cos\alpha}$

Applying half-angle identities:
$\sin x = \frac{2\sin^2(\frac{\alpha}{2})}{2\cos^2(\frac{\alpha}{2})} = \frac{\sin^2(\frac{\alpha}{2})}{\cos^2(\frac{\alpha}{2})} = \tan^2\left(\frac{\alpha}{2}\right)$

The final answer is $\boxed{\text{D}}$