1. Given limit

We need to evaluate

$$L=\lim_{x\to \frac\pi2}\tan^2 x\left(\sqrt{2\sin^2 x+3\sin x+4}-\sqrt{\sin^2 x+6\sin x+2}\right).$$

As $x\to \frac\pi2$, we have $\sin x\to 1$ and $\tan^2 x\to \infty$, while the bracket tends to

$$\sqrt{2+3+4}-\sqrt{1+6+2}=3-3=0.$$

So this is of the indeterminate form $\infty\cdot 0$.

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2. Rationalize the expression inside the bracket

Let

$$A=2\sin^2 x+3\sin x+4,\qquad B=\sin^2 x+6\sin x+2.$$

Then

$$\sqrt A-\sqrt B=\frac{A-B}{\sqrt A+\sqrt B}.$$

Hence

$$L=\lim_{x\to \frac\pi2}\tan^2 x\cdot \frac{A-B}{\sqrt A+\sqrt B}.$$

Now,

$$A-B=(2\sin^2 x+3\sin x+4)-(\sin^2 x+6\sin x+2)$$ $$=\sin^2 x-3\sin x+2.$$

Factorize:

$$\sin^2 x-3\sin x+2=(\sin x-1)(\sin x-2).$$

So,

$$L=\lim_{x\to \frac\pi2}\tan^2 x\cdot \frac{(\sin x-1)(\sin x-2)}{\sqrt A+\sqrt B}.$$

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3. Use identities near $x=\frac\pi2$

Recall

$$\tan^2 x=\frac{\sin^2 x}{\cos^2 x},$$

and

$$1-\sin^2 x=\cos^2 x \implies (1-\sin x)(1+\sin x)=\cos^2 x.$$

Thus

$$\sin x-1=-(1-\sin x)=-\frac{\cos^2 x}{1+\sin x}.$$

Substitute this:

$$L=\lim_{x\to \frac\pi2}\frac{\sin^2 x}{\cos^2 x}\cdot \frac{\left(-\frac{\cos^2 x}{1+\sin x}\right)(\sin x-2)}{\sqrt A+\sqrt B}.$$

The $\cos^2 x$ cancels:

$$L=\lim_{x\to \frac\pi2}-\frac{\sin^2 x(\sin x-2)}{(1+\sin x)(\sqrt A+\sqrt B)}.$$

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4. Now substitute $\sin x\to 1$

As $x\to \frac\pi2$:

• $\sin x\to 1$
• $\sin^2 x\to 1$
• $\sin x-2\to -1$
• $1+\sin x\to 2$
• $A\to 9$, so $\sqrt A\to 3$
• $B\to 9$, so $\sqrt B\to 3$

Therefore,

$$L=-\frac{1\cdot(-1)}{2(3+3)}=\frac{1}{12}.$$

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5. Option check

Thus,

$$\boxed{\frac{1}{12}}$$

which corresponds to Option A.

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6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.