Let $α = Tan (\frac{5 π}{16} Sin (2 {Cos}^{- 1} (\frac{1}{\sqrt{5}}))) and β = Cos ({Sin}^{- 1} (\frac{4}{5}) + {Sec}^{- 1} (\frac{5}{3}))$ Where the inverse Trigonometric functions take principal values then The equation whose roots are $α$ and $β$ is

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$25 x^{2} - 32 x + 7 = 0$

$5 x^{2} - 12 x + 7 = 0$

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

$15 x^{2} - 8 x - 7 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} Cos θ = \frac{1}{\sqrt{5}}, Sin θ = \frac{2}{\sqrt{5}}, Sin 2 θ = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5} \\ α = Tan (\frac{5 π}{16} \cdot \frac{4}{5}) = Tan (\frac{π}{4}) = 1 \end{array}$
$\begin{array}{l} β = Cos ({Sin}^{- 1} (\frac{4}{5}) + {Cos}^{- 1} \frac{3}{5}) = Cos ({Cos}^{- 1} (\frac{3}{5}) + {Cos}^{- 1} (\frac{3}{5})) \\ = Cos (2 {Cos}^{- 1} (\frac{3}{5})) = Cos 2 θ, Cos θ = \frac{3}{5} \\ = 2 {Cos}^{2} θ - 1 = 2 \cdot (\frac{9}{25}) - 1 \\ β = \frac{- 7}{25} \end{array}$