Let $x, y$ be real numbers such that $x \neq y$ and $x y \neq 1$. If $a x+b \sec \left(\tan ^{-1}…
Let $x, y$ be real numbers such that $x \neq y$ and $x y \neq 1$. If $a x+b \sec \left(\tan ^{-1} x\right)=c$ and $a y+b \sec \left(\tan ^{-1} y\right)=c$, then $\frac{x+y}{1-x y}=$
$\frac{2 a b}{a^2-b^2}$
$\frac{2 a c}{a^2+c^2}$
$\frac{2 a b}{a^2+b^2}$
$\frac{2 a c}{a^2-c^2}$ $\tanh ^{-1} \frac{1}
Solution
Let $x=\tan \theta$, then we have
$\begin{aligned}
& & a \tan \theta+b \sec \theta & =c \\
\Rightarrow & & a \sin \theta+b & =c \cos \theta \\
\Rightarrow & & c \cos \theta-a \sin b & =b \\
\text {Let } & & a^2+c^2 & =r^2
\end{aligned}$
Let
So, for some $\alpha$,
$\begin{aligned}
& c=r \cos \alpha, \\
& a=r \sin \alpha
\end{aligned}$
Also, $\tan \alpha=\frac{a}{c}$
Thus, $\cos \alpha \cos \theta-\sin \alpha \sin \theta=\frac{b}{r}$
$\Rightarrow \cos (\alpha+\theta)=\frac{b}{r} \Rightarrow \alpha+\theta= \pm \cos ^{-1} \frac{b}{r}$
Let $\alpha+\theta$ be the positive solution and $\alpha+\phi$ the negative solution, where
$\begin{aligned}
& y=\tan \phi \\
& \therefore \alpha+\phi=-(\alpha+\theta) \\
& \Rightarrow -2 \alpha=\theta+\phi \\
& \Rightarrow \tan (-2 x)=\tan (\theta+\phi) \\
& \Rightarrow \frac{-2 \tan \alpha}{1-\tan ^2 \alpha}=\tan (\theta+\phi) \\
& \Rightarrow \frac{-2 a / c}{1-a^2 / c^2}=\frac{x+y}{1-x y} \Rightarrow \frac{2 a c}{a^2-c^2}=\frac{x+y}{1-x y} . \\
\end{aligned}$