If $U = cot^{-1} \sqrt{cos 2\theta } -tan^{-1}\sqrt{cos 2\theta }$, then sin U equals

$tan^2\theta $

We have,

 $U = cot^{-1} \sqrt{cos 2\theta } -tan^{-1}\sqrt{cos 2\theta }$

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

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

$⇒ U = \frac{\pi}{2} - tan^{-1} \left(\frac{2\sqrt{cos 2\theta }}{1-cos2 \theta}\right)$

$⇒ U = \frac{\pi}{2} - tan^{-1} \left(\frac{\sqrt{cos 2\theta }}{sin^2 \theta}\right)$

$⇒ tan \left(\frac{\pi}{2} -U\right) = \frac{\sqrt{cos 2\theta }}{sin^2 \theta}$

$⇒ cot U = \frac{\sqrt{cos 2\theta }}{sin^2 \theta}$

$⇒ sin U =\frac{1}{1+cot^2 \theta }=\frac{1}{1+\frac{cos2\theta}{sin^4\theta}}$

$⇒ sin U =\frac{sin^2\theta}{\sqrt{sin^4\theta + 1 - 2 sin^2 \theta }}$

$⇒ sin U =\frac{sin^2\theta}{|1-sin^2\theta|}= tan^2 \theta $