Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$x=1:y=1$ is not a solution of the given equation

Suppose $(x,y)\ne (1,1)$

Then $\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{xy}}=\frac{1}{z}\Rightarrow x(z-y)=-(1+yz)\Rightarrow x=-\frac{(1+yz)}{z-y}$

If $y=n+1:z=n$ then $x=n^2+n+1$

All such numbers are solutions