If $a x + b (\sec (\tan^{- 1} x)) = c$ and $a y + b (\sec (\tan^{- 1} y)) = c$

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$x + y = \frac{2 a c}{a^{2} - b^{2}}$

$x y = \frac{a^{2} - b^{2}}{a^{2} - c^{2}}$

$x y = \frac{c^{2} - b^{2}}{a^{2} - b^{2}}$

$x + y = \frac{2 a b}{a^{2} - b^{2}}$

answer is B, C.

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Detailed Solution

Given $a x + b (\sec (\tan^{- 1} x)) = c$ and $a y + b (\sec (\tan^{- 1} y)) = c$
Let $\tan^{- 1} x = α$ and $\tan^{- 1} y = β,$ then the given relations are
$a \tan α + b \sec α = c$ and $a \tan β + b \sec β = c$
From these two relations, we can conclude that equation
$a \tan θ + b \sec θ = c$ has roots $α$ and $β$
$a \tan θ + b \sec θ = c$
Or $b \sec θ = c - a \tan θ$
Or $b^{2} \sec^{2} θ = c^{2} - 2 a c \tan θ + a^{2} \tan^{2} θ$
Or $b^{2} + b^{2} \tan^{2} θ = c^{2} - 2 a c \tan θ + a^{2} \tan^{2} θ$
Or $(a^{2} - b^{2}) \tan^{2} θ - 2 a c \tan θ + c^{2} - b^{2} = 0$
Therefore, sum of the roots, $\tan α + \tan β = x + y = \frac{2 a c}{a^{2} - b^{2}}$
And the product of roots, $\tan α \tan β = x y = \frac{c^{2} - b^{2}}{a^{2} - b^{2}}$