Question

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

• A. $\frac{3}{4}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$
• D. $\frac{5}{12}$

Answer 1

The correct option is C $\frac{1}{12}$

${lim}_{x\to \frac{\pi }{2}}\frac{\sqrt{2\sin {}^{2}x+3\sin x+4}-\sqrt{\sin {}^{2}x+6\sin x+2}}{\cot {}^{2}x}$

Using LHR.

$\Rightarrow \frac{\frac{4\sin x\cos x+3\cos x}{2\sqrt{2\sin {}^{2}x+3\sin x+4}}-\frac{(2\sin x\cos x+6\cos x)}{2\sqrt{\sin {}^{2}x+6\sin x+2}}}{2\cot x-\csc {}^{2}x}\Rightarrow \frac{\cos x[\frac{4\sin x+3}{2\sqrt{2\sin {}^{2}x+3\sin x+4}}-\frac{(2\sin x+6)}{2\sqrt{\sin {}^{2}x+6\sin x+2}}]}{-2\frac{\cos x}{\sin {}^{3}x}}$

Cancelling cosx from numerator and denominator.

$\Rightarrow \frac{\frac{4+3}{2\sqrt{9}}-\frac{2+6}{2\sqrt{9}}}{\frac{-2}{1}}$

$\Rightarrow \frac{\frac{1}{2\sqrt{9}}(7-8)}{-2}\Rightarrow \frac{1}{4\sqrt{9}}\Rightarrow \frac{1}{12}$