If `ax + b sec(tan^-1 x) = c and ay + b sec(tan^-ly) = c,` then `(x+y)/(1-xy)` is equal to

Options

A. `(2ab)/(a^(2) -b^(2))`
B. `(c^(2) -b^(2))/(a^(2) -b^(2))`
C. `(c^(2) -b^(2))/(a^(2) + b^(2))`
D. none of these

Answer

B

Explanation

Given `ax + b (sec(tan^(-1)x)) = c and ay + b(sec(tan^(-1)y)) = c`
Let `tan^(-1) x = alpha and tan^(-1) y = beta`, then the given relations are `a tan alpha + b sec alpha = c and a tan beta + b sec beta = c`
From these two releations, we can conclude that equation
`a tan theta + b sec theta = c` has roots `alpha and beta`
`a tan theta and b sec theta = c`
or `b sec theta = c - a tan theta`
or `b^(2) sec^(2) theta = c^(2) -2 ac tan theta + a^(2) tan^(2) theta`
or `b^(2) + b^(2) tan^(2) theta = c^(2) -2 ac tan theta + a^(2) tan^(2) theta`
or `(a^(2) -b^(2)) tan^(2) theta - 2 ac tan theta + c^(2) -b^(2) = 0`
Therefore, sum of the roots, `tan alpha + beta = xy = (c^(2) -b^(2))/(a^(2) -b^(2)) and (x + y)/(1-xy) = ((2ac)/(a^(2) -b^(2)))/(1-(c^(2) -b^(2))/(a^(2) -b^(2))) = (2ac)/(a^(2) -c^(2))`