x,y∈R∋x≠y and x≠1. If ax+bsec(tan-1x)=c and ay+bsec(tan-1y)=c then x+y1-xy=

ax+bsec⁡tan−1⁡x=c⇒ax+bsec⁡sec−1⁡1+x2=c⇒ax+b1+x2=c⇒b1+x2=c−ax⇒b21+x2=c2+a2x2−2acx⇒b2−a2x2+b2−c2+2acx=0⇒x=−2ac±4a2c2−4b2−a2b2−c22b2−a2similarly,ay+bsec⁡tan−1⁡y=c⇒y=−2ac±4a2c2−4b2−a2b2−c22b2−a2 xy=−2ac+4a2c2−4b2−a2b2−c22b2−a2×−2ac−4a2c2−4b2−a2b2−c22b2−a2=4a2c2−4a2c2−4b2−a2b2−c24b2−a22=c2−b2a2−b2x+y=2aca2−b2x+y1−xy=2aca2−c2

x,y∈R∋x≠y and x≠1. If ax+bsec(tan-1x)=c and ay+bsec(tan-1y)=c then x+y1-xy=

ax+bsec⁡tan−1⁡x=c⇒ax+bsec⁡sec−1⁡1+x2=c⇒ax+b1+x2=c⇒b1+x2=c−ax⇒b21+x2=c2+a2x2−2acx⇒b2−a2x2+b2−c2+2acx=0⇒x=−2ac±4a2c2−4b2−a2b2−c22b2−a2similarly,ay+bsec⁡tan−1⁡y=c⇒y=−2ac±4a2c2−4b2−a2b2−c22b2−a2 xy=−2ac+4a2c2−4b2−a2b2−c22b2−a2×−2ac−4a2c2−4b2−a2b2−c22b2−a2=4a2c2−4a2c2−4b2−a2b2−c24b2−a22=c2−b2a2−b2x+y=2aca2−b2x+y1−xy=2aca2−c2

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$x, y \in R ∋ x \neq y$ and $x \neq 1 .$ If $ax + bsec (\tan^{- 1} x) = c$ and $ay + bsec (\tan^{- 1} y) = c then \frac{x + y}{1 - xy} =$

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$\frac{2 ab}{a^{2} - b^{2}}$

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

$\frac{2 ab}{a^{2} + b^{2}}$

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

answer is D.

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Detailed Solution

$\begin{array}{l} a x + b (\sec (\tan^{- 1} x)) = c \\ \Rightarrow ax + b (\sec (\sec^{- 1} \sqrt{1 + x^{2}})) = c \\ \Rightarrow ax + b \sqrt{1 + x^{2}} = c \Rightarrow b \sqrt{1 + x^{2}} = c - ax \\ \Rightarrow b^{2} (1 + x^{2}) = c^{2} + a^{2} x^{2} - 2 acx \\ \Rightarrow (b^{2} - a^{2}) x^{2} + (b^{2} - c^{2}) + 2 acx = 0 \\ \Rightarrow x = \frac{- 2 ac \pm \sqrt{4 a^{2} c^{2} - 4 (b^{2} - a^{2}) (b^{2} - c^{2})}}{2 (b^{2} - a^{2})} \\ s i m i l a r l y, \\ \begin{array}{l} a y + b \sec (\tan^{- 1} y) = c \\ \Rightarrow y = \frac{- 2 ac \pm \sqrt{4 a^{2} c^{2} - 4 (b^{2} - a^{2}) (b^{2} - c^{2})}}{2 (b^{2} - a^{2})} \begin{array}{l} x y = \frac{- 2 a c + \sqrt{4 a^{2} c^{2} - 4 (b^{2} - a^{2}) (b^{2} - c^{2})}}{2 (b^{2} - a^{2})} \times \frac{- 2 a c - \sqrt{4 a^{2} c^{2} - 4 (b^{2} - a^{2}) (b^{2} - c^{2})}}{2 (b^{2} - a^{2})} \\ = \frac{4 a^{2} c^{2} - 4 a^{2} c^{2} - 4 (b^{2} - a^{2}) (b^{2} - c^{2})}{4 {(b^{2} - a^{2})}^{2}} = \frac{c^{2} - b^{2}}{a^{2} - b^{2}} \\ x + y = \frac{2 a c}{(a^{2} - b^{2})} \\ \frac{x + y}{1 - x y} = \frac{2 a c}{a^{2} - c^{2}} \end{array} \end{array} \end{array}$