QJEEM22Mabc85: JEE Main 2022 Mathematics - Class 12 JEEM22Mabc85 Solution: Let and where the inverse trigonometric functions take…

Given,

$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$

We know, $2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$

$\therefore$ $2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times {1 \over 5} - 1} \right) = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right)$

$\therefore$ $\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( { - {3 \over 5}} \right)} \right.} \right)$

$= \tan \left( {{{5\pi } \over {16}}\sin \left( {\pi - {{\cos }^{ - 1}}\left( { {3 \over 5}} \right)} \right.} \right)$

$= \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right.} \right)$

$= \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right)} \right.} \right)$

$= \tan \left( {{{5\pi } \over {16}} \times {4 \over 5}} \right)$

$= \tan \left( {{\pi \over 4}} \right)$

$= 1$

$\therefore$ $\alpha = 1$

Also given,

$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$

$= \cos \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$

$= \cos \left( {2{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$

$= \cos \left( {{{\cos }^{ - 1}}\left( {2\left. {{{\left( {{3 \over 5}} \right)}^2} - 1} \right)} \right.} \right)$

$= \cos \left( {{{\cos }^{-1}}\left( {{{18} \over {25}} - 1} \right.} \right)$

$= {{18} \over {25}} - 1$

$= {{18 - 25} \over {25}}$

$= - {7 \over {25}}$

$\therefore$ $\beta = - {7 \over {25}}$

$\therefore$ The quadratic equation with roots $\alpha$ and $\beta$ is

${x^2} - (\alpha + \beta )x + \alpha \beta = 0$

$\Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0$

$\Rightarrow 25{x^2} - 18x - 7 = 0$