If $1+\sqrt{1+\mathrm{a}}=(1+\sqrt{1-\mathrm{a}})\cot \mathrm{\alpha }$ and $0<\mathbf{a}<1,$then $\sin 4\mathrm{\alpha }=$

Let $a=\sin 4\theta \Rightarrow \sqrt{1+a}=\cos 2\theta +\sin 2\theta$ and

$\begin{array}{l}\sqrt{1-a}=\cos 2\theta -\sin 2\theta \end{array}$

$\left(1+\sqrt{1+\mathrm{a}}\right)=(1+\sqrt{1-\mathrm{a}})\cot \mathrm{\alpha }$

$\left(1+\cos 2\theta +\sin 2\theta \right)=\left(1+\cos 2\theta -\sin 2\theta \right)cot\alpha$

$\left(2{\cos }^2\theta +2\sin \theta \cos \theta \right)=\left(2{\cos }^2\theta -2\sin \theta \cos \theta \right)cot\alpha$

$\begin{array}{l}\Rightarrow \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }=\cot \alpha \Rightarrow \frac{1+\tan \theta }{1-\tan \theta }=\cot \alpha \\ \Rightarrow \tan \left(\frac{\pi }{4}+\theta \right)=\tan \left(\frac{\pi }{2}-\alpha \right)\\ \Rightarrow \theta =\left(\frac{\pi }{4}-\alpha \right)\\ \Rightarrow a=\sin 4\theta =\sin (\pi -4\alpha )=\sin 4\alpha \end{array}$