If $ u = cot^{-1}\sqrt{cos \theta } - tan^{-1} \sqrt{cos \theta }$, then sin u = 

$\frac{tan^2\theta }{2}$

We have,

$ u = cot^{-1}\sqrt{cos \theta} - tan^{-1} \sqrt{cos \theta }$

$⇒ u = \frac{\pi}{2} - 2tan^{-1}\sqrt{cos\theta }$

$⇒ u = \frac{\pi}{2} - 2 \alpha , $ where $ \alpha = tan^{-1}(\sqrt{cos \theta})$

$⇒ sin u =cos 2 \alpha $

$⇒ sin  u = \frac{1-tan^2 \alpha}{1+tan^2 \alpha}=\frac{1-cos \theta}{1+cos \theta}=tan^2 \frac{\theta }{2}$