Let x, y be real numbers such that x neq y an | vedclass

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(D) We know that $\sec(	an^{-1} x) = \sqrt{1+x^2}$. Given the equations $ax + b\sqrt{1+x^2} = c$ and $ay + b\sqrt{1+y^2} = c$,we can rewrite them as

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

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(D) We know that $\sec(	an^{-1} x) = \sqrt{1+x^2}$. Given the equations $ax + b\sqrt{1+x^2} = c$ and $ay + b\sqrt{1+y^2} = c$,we can rewrite them as $b\sqrt{1+x^2} = c - ax$ and $b\sqrt{1+y^2} = c - ay$.Squaring both sides,we get $b^2(1+x^2) = c^2 - 2acx + a^2x^2$ and $b^2(1+y^2) = c^2 - 2acy + a^2y^2$.Rearranging these,we have $(a^2-b^2)x^2 - 2acx + (c^2-b^2) = 0$ and $(a^2-b^2)y^2 - 2acy + (c^2-b^2) = 0$.This implies that $x$ and $y$ are the distinct roots of the quadratic equation $(a^2-b^2)t^2 - 2act + (c^2-b^2) = 0$.Using the properties of roots,the sum of roots $x+y = \frac{2ac}{a^2-b^2}$ and the product of roots $xy = \frac{c^2-b^2}{a^2-b^2}$.Now,calculate $1-xy = 1 - \frac{c^2-b^2}{a^2-b^2} = \frac{a^2-b^2-c^2+b^2}{a^2-b^2} = \frac{a^2-c^2}{a^2-b^2}$.Therefore,$\frac{x+y}{1-xy} = \frac{2ac / (a^2-b^2)}{(a^2-c^2) / (a^2-b^2)} = \frac{2ac}{a^2-c^2}$.

Let $x, y$ be real numbers such that $x \neq y$ and $xy \neq 1$. 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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