If tan−1x+tan−1y+tan−1z=π, then x+y+z=
xyz
0
1
2xyz
Explanation for the correct option:
Use formula tan−1A−tan−1B=tan−1A−B1+AB
Given equation tan−1x+tan−1y+tan−1z=π, then
tan−1x+tan−1y=π−tan−1ztan−1x+y1−xy=(π−tan−1z)x+y1−xy=tan(tan−1(−z))x+y1−xy=−zx+y=−z(1−xy)x+y=−z+xyzx+y+z=xyz
Therefore, x+y+z=xyz.
Hence, the correct option is (A).
If x + y + z = xyz, then tan-1x + tan-1y + tan-1z =