Let $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$ and $$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$ where the inverse trigonometric functions take principal values. Then, the equation whose roots are $\alpha$ and $\beta$ is :

<p>Given,</p> <p>$$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$</p> <p>We know, $2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$</p> <p>$\therefore$ $$2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times {1 \over 5} - 1} \right) = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right)$$</p> <p>$\therefore$ $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( { - {3 \over 5}} \right)} \right.} \right)$$</p> <p>$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {\pi - {{\cos }^{ - 1}}\left( { {3 \over 5}} \right)} \right.} \right)$$</p> <p>$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right.} \right)$$</p> <p>$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right)} \right.} \right)$$</p> <p>$= \tan \left( {{{5\pi } \over {16}} \times {4 \over 5}} \right)$</p> <p>$= \tan \left( {{\pi \over 4}} \right)$</p> <p>$= 1$</p> <p>$\therefore$ $\alpha = 1$</p> <p>Also given,</p> <p>$$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$</p> <p>$$ = \cos \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$$</p> <p>$= \cos \left( {2{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$</p> <p>$$ = \cos \left( {{{\cos }^{ - 1}}\left( {2\left. {{{\left( {{3 \over 5}} \right)}^2} - 1} \right)} \right.} \right)$$</p> <p>$= \cos \left( {{{\cos }^{-1}}\left( {{{18} \over {25}} - 1} \right.} \right)$</p> <p>$= {{18} \over {25}} - 1$</p> <p>$= {{18 - 25} \over {25}}$</p> <p>$= - {7 \over {25}}$</p> <p>$\therefore$ $\beta = - {7 \over {25}}$</p> <p>$\therefore$ The quadratic equation with roots $\alpha$ and $\beta$ is</p> <p>${x^2} - (\alpha + \beta )x + \alpha \beta = 0$</p> <p>$$ \Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0$$</p> <p>$\Rightarrow 25{x^2} - 18x - 7 = 0$</p>