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

Given,

$$\lim_{x\to\frac{\pi}{2}}\tan^2x\left(\sqrt{2\sin^2x+3\sin x+4}-\sqrt{\sin^2x+6\sin x+2}\right)$$

Let

$$u=\sin x$$

As

$$x\to\frac{\pi}{2}$$

we have

$$u\to1$$

Also,

$$\tan^2x=\frac{\sin^2x}{1-\sin^2x}=\frac{u^2}{1-u^2}$$

Hence, the limit becomes

$$\lim_{u\to1}\frac{u^2}{1-u^2}\left(\sqrt{2u^2+3u+4}-\sqrt{u^2+6u+2}\right)$$

Rationalizing,

$$=\lim_{u\to1}\frac{u^2\left[(2u^2+3u+4)-(u^2+6u+2)\right]}{(1-u^2)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

$$=\lim_{u\to1}\frac{u^2(u^2-3u+2)}{(1-u)(1+u)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

$$=\lim_{u\to1}\frac{u^2(u-1)(u-2)}{(1-u)(1+u)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

Since,

$$1-u=-(u-1)$$

we get

$$=\lim_{u\to1}\frac{-u^2(u-2)}{(1+u)\left(\sqrt{2u^2+3u+4}+\sqrt{u^2+6u+2}\right)}$$

Substituting

$$u=1$$

$$=\frac{-1(1-2)}{2(3+3)}$$

$$=\frac1{12}$$

Hence, $$\boxed{\frac1{12}}$$.