If $\tan x/2=\mathrm{cosec}x-\sin x$, then ${\sec }^2(x/2)=$

$\tan (x/2)=\frac{1+{\tan }^2(x/2)}{2\tan (x/2)}-\frac{2\tan (x/2)}{1+{\tan }^2(x/2)}$

$\begin{array}{l}\Rightarrow 2{\tan }^2(x/2)\left(1+{\tan }^2(x/2)={\left[1+{\tan }^2(x/2)\right]}^2-4{\tan }^2(x/2)\right.\\ \Rightarrow 2{\tan }^4(x/2)+2{\tan }^2(x/2)=1+{\tan }^4(x/2)-2{\tan }^2(x/2)\\ \Rightarrow {\tan }^4(x/2)+4{\tan }^2(x/2)-1=0\\ \Rightarrow {\tan }^2(x/2)=\sqrt{5}-2\Rightarrow {\sec }^2(x/2)=\sqrt{5}-1\end{array}$