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 ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$ $\Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0$ $\Rightarrow 25{x^2} - 18x - 7 = 0$