Q.If tan^-1 x + tan^-1 y + tan^-1 z = pi/2 and x + y + z = sqrt(3), then show that x = y = z.

Convert the arctan sum condition into $xy+yz+zx=1$, then use it with $x+y+z=\sqrt3$ to show $(x-y)^2+(y-z)^2+(z-x)^2=0$.

Step 1 — get $xy+yz+zx=1$. From $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi/2$:

$\tan^{-1}x+\tan^{-1}y=\dfrac{\pi}{2}-\tan^{-1}z=\cot^{-1}z=\tan^{-1}\dfrac1z$

Taking tangent of both sides (using the addition formula):

$\dfrac{x+y}{1-xy}=\dfrac1z \;\Rightarrow\; z(x+y)=1-xy \;\Rightarrow\; xy+yz+zx=1$

Step 2 — use the sum of squares identity. We're given $x+y+z=\sqrt3$. Then

$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=3-2(1)=1$

Now consider …