If ax+bsec⁡tan−1⁡x=c and ay+bsec⁡tan−1⁡y=c, then x+y1−xy=

Let tan−1⁡x=α and tan−1⁡y=β⇒tan⁡α=x,tan⁡β=yThe given system of equations isatan⁡α+bsec⁡α=c and atan⁡β+bsec⁡β=c∴α and β are roots of a tan⁡θ+bsec⁡θ=c⇒(bsec⁡θ)2=(c−atan⁡θ)2⇒a2−b2tan2⁡θ−2actan⁡θ+c2−b2=0∴tan⁡α+tan⁡β=2aca2−b2 and tan⁡αtan⁡β=c2b2a2−b2∴x+y=c2−b2a2−b2 and xy=c2−b2a2−b2⇒ 1−xy=c2−b2a2−b2∴ x+y1−xy=2aca2−c2

If ax+bsec⁡tan−1⁡x=c and ay+bsec⁡tan−1⁡y=c, then x+y1−xy=

Let tan−1⁡x=α and tan−1⁡y=β⇒tan⁡α=x,tan⁡β=yThe given system of equations isatan⁡α+bsec⁡α=c and atan⁡β+bsec⁡β=c∴α and β are roots of a tan⁡θ+bsec⁡θ=c⇒(bsec⁡θ)2=(c−atan⁡θ)2⇒a2−b2tan2⁡θ−2actan⁡θ+c2−b2=0∴tan⁡α+tan⁡β=2aca2−b2 and tan⁡αtan⁡β=c2b2a2−b2∴x+y=c2−b2a2−b2 and xy=c2−b2a2−b2⇒ 1−xy=c2−b2a2−b2∴ x+y1−xy=2aca2−c2

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If $ax + b (\sec (\tan^{- 1} x)) = c$ and $ay + b (\sec (\tan^{- 1} y)) = c$ , then $\frac{x + y}{1 - xy} =$

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$\frac{2 ac}{a^{2} - c^{2}}$

$\frac{2 ac}{a^{2} + c^{2}}$

$\frac{ac}{a^{2} + c^{2}}$

$\frac{ac}{a^{2} - c^{2}}$

answer is B.

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Detailed Solution

Let $\tan^{- 1} x = α and \tan^{- 1} y = β$

$\Rightarrow \tan α = x, \tan β = y$

The given system of equations is

$atan α + bsec α = c$ and $atan β + bsec β = c$

$∴ α and β are roots of a \tan θ + bsec θ = c$

$\begin{array}{l} \Rightarrow (bsec θ)^{2} = (c - atan θ)^{2} \\ \Rightarrow (a^{2} - b^{2}) \tan^{2} θ - 2 actan θ + c^{2} - b^{2} = 0 \\ ∴ \tan α + \tan β = \frac{2 ac}{a^{2} - b^{2}} and \tan αtan β = \frac{c^{2} b^{2}}{a^{2} - b^{2}} \\ ∴ x + y = \frac{c^{2} - b^{2}}{a^{2} - b^{2}} and xy = \frac{c^{2} - b^{2}}{a^{2} - b^{2}} \\ \Rightarrow 1 - xy = \frac{c^{2} - b^{2}}{a^{2} - b^{2}} \\ ∴ \frac{x + y}{1 - xy} = \frac{2 ac}{a^{2} - c^{2}} \end{array}$