ax + bsec(tan^-1(x)) = c
ay + bsec(tan^-1(y)) = c , then prove that (x+y)/(1-xy) = 2ac/(a^2-c^2)

Dear student,
given $ax+bsec\left({\tan }^{-1}x\right)=c$   ............(1)
since ${\tan }^{-1}x=se{c}^{-1}\sqrt{1+x^2}$
now equation (1) will become
$ax+bsec\left(se{c}^{-1}\sqrt{1+x^2}\right)=c$
we know that $sec\left(se{c}^{-1}x\right)=x$
$\Rightarrow ax+b\sqrt{1+x^2}=c$      .............(2)
similar as above , equation $ay+bsec\left({\tan }^{-1}y\right)=c$ will become
$\Rightarrow ay+b\sqrt{1+y^2}=c$   ............(3)
now from (2) $b=\frac{c-ax}{\sqrt{1+x^2}}$ and from (3) $b=\frac{c-ay}{\sqrt{1+y^2}}$
equating both values of b we have
$\frac{c-ax}{\sqrt{1+x^2}}=\frac{c-ay}{\sqrt{1+y^2}}$
squaring both sides we get
$\frac{c^2+a^2{x}^2-2acx}{1+x^2}=\frac{c^2+a^2{y}^2-2acy}{1+y^2}$
cross multiplying'
$c^2+c^2{y}^2+a^2{x}^2+a^2{x}^2{y}^2-2acx-2acx{y}^2=c^2+c^2{x}^2+a^2{y}^2+a^2{x}^2{y}^2-2acy-2acy{x}^2$
$c^2(y^2-x^2)+a^2(x^2-y^2)+2ac(y-x)+2acxy(x-y)=0$
or $a^2(x+y)-c^2(x+y)=2ac-2acxy$
or $(a^2-c^2)(x+y)=2ac(1-xy)$
or $\frac{x+y}{1-xy}=\frac{2ac}{a^2-c^2}$   hence proved

enjoy