QJEEM22Mae34b: JEE Main 2022 Mathematics - Class 12 JEEM22Mae34b Solution

$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$

Let ${\cos ^{ - 1}}x = t$

$\Rightarrow x = \cos t$

When $x \to {1 \over {\sqrt 2 }}$ , then $t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}$

$\therefore$ $\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}$

$= \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}$

$= \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}$

$= \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t$

$= - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t$

$= - {1 \over {\sqrt 2 }}$