QJEEM22Mad8e2: JEE Main 2022 Mathematics - Class 12 JEEM22Mad8e2 Solution

$\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}$

$= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }^2}x + 6\sin x + 2} }}$

$= {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(1 - \sin x)(2 - \sin x)} \over {{{\cos }^2}x}}\,.\,{\sin ^2}x$

$= {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(2 - \sin x){{\sin }^2}x} \over {1 + \sin x}}$

$= {1 \over {12}}$