Question
If $$ax + b\left( {\sec \left( {{{\tan }^{ - 1}}x} \right)} \right) = c\,$$ and $$ay + b\left( {\sec \left( {{{\tan }^{ - 1}}y} \right)} \right) = c,\,$$ then $$\frac{{x + y}}{{1 - xy}} = $$
A.
$$\frac{{ac}}{{{a^2} + {c^2}}}$$
B.
$$\frac{{2ac}}{{{a} - {c}}}$$
C.
$$\frac{{2ac}}{{{a^2} - {c^2}}}$$
D.
$$\frac{{a + c}}{{{1} - {ac}}}$$
Answer :
$$\frac{{2ac}}{{{a^2} - {c^2}}}$$
Solution :
$$\eqalign{
& {\text{Let, }}{\tan ^{ - 1}}x = \alpha {\text{ and }}{\tan ^{ - 1}}y = \beta \cr
& \Rightarrow \tan \alpha = x,\tan \beta = y \cr} $$
The given system of equations is
$$a\tan \alpha + b\sec \alpha = c{\text{ and }}a\tan \beta + b\sec \beta = c$$
$$\therefore \alpha {\text{ and }}\beta $$ are the roots of $$a\tan \theta + b\sec \theta = c$$
$$\eqalign{
& \Rightarrow {\left( {b\sec \theta } \right)^2} = {\left( {c - a\tan \theta } \right)^2} \cr
& \Rightarrow \left( {{a^2} - {b^2}} \right){\tan ^2}\theta - 2ac\tan \theta + {c^2} - {b^2} = 0 \cr
& \Rightarrow \tan \alpha + \tan \beta = \frac{{2ac}}{{{a^2} - {b^2}}}{\text{ and }}\tan \alpha \tan \beta = \frac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}} \cr
& \Rightarrow x + y = \frac{{2ac}}{{{a^2} - {b^2}}}{\text{ and }}xy = \frac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}} \cr
& \Rightarrow 1 - xy = \frac{{{a^2} - {c^2}}}{{{a^2} - {b^2}}} \cr
& \Rightarrow \frac{{x + y}}{{1 - xy}} = \frac{{2ac}}{{{a^2} - {c^2}}} \cr} $$