JEE Math Practice Question
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$$ is equal to
- A.$${1 \over {12}}$$
- B.$$-$$$${1 \over {18}}$$
- C.$$-$$$${1 \over {12}}$$
- D.$${1 \over {6}}$$
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