If tan−1x+tan−1y+tan−1z=π, then x+y+z is equal to
xyz
0
1
2xyz
We have,
tan−1x+tan−1y+tan−1z=tan−1x+y+z−xyz1−xy−yz−zx
⇒ π=tan−1x+y+z−xyz1−(xy+yz+zx)⇒ x+y+z=xyz