Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The point $x=1,y=1$ is not a solution to the given equation.

Suppose $(x,y)\ne (1,1)$. Then, we can simplify the equation as follows:

$$\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{xy}}=\frac{1}{z}$$

This implies:

$$x(z-y)=-(1+yz)$$

Solving for $x$, we get:

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

If we set $y=n+1$ and $z=n$, then the value of $x$ becomes:

$$x=n^2+n+1$$

All such numbers derived from this relationship are solutions to the equation.