$\underset{x\to \frac{\pi }{2}}{\lim }\left({\tan }^{2}x\left({\left(2{\sin }^{2}x+3\sin x+4\right)}^{\frac{1}{2}}-{\left({\sin }^{2}x+6\sin x+2\right)}^{\frac{1}{2}}\right)\right)$ is equal to 

$\underset{x\to \frac{\pi }{2}}{\lim }{\tan }^{2}x\left[\sqrt{2{\sin }^{2}x+\sin x+4}\right]-\sqrt{{\sin }^{2}x+6\sin x+2}$

Rationalize the functions apply the limit in the denominator.

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\tan }^{2}x\left[{\sin }^{2}x-3\sin x+2\right]}{\sqrt{9}+\sqrt{9}}=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\tan }^{2}x\left(\sin x-1\right)\left(\sin x-2\right)}{6}$

$=\frac{1}{6}\underset{x\to \frac{\pi }{2}}{\lim }{\tan }^{2}x\left(1-\sin x\right)=\frac{1}{6}\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\sin }^{2}x\left(1-\sin x\right)}{\left(1-\sin x\right)\left(1+\sin x\right)}=\frac{1}{12}$