Question

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{\pi}{2}$, then ________.

• A. $x+y+z-xyz=0$
• B. $xy+yz+zx-1=0$
• C. $x+y+z+xyz=0$
• D. $xy+yz+zx+1=0$

Hint:

If $\sum \tan^{-1}x = \pi/2$, then $\sum xy = 1$. If $\sum \tan^{-1}x = \pi$, then $\sum x = xyz$.

Answer 1

The Correct Option is B

Step 1: Concept
Formula: $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1} \left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right)$.
Step 2: Analysis
Given $\tan^{-1} \left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right) = \frac{\pi}{2}$.
Step 3: Calculation
The tangent of $\pi/2$ is undefined (approaches infinity), which means the denominator of the fraction must be zero.
$1 - xy - yz - zx = 0$.
$xy + yz + zx = 1$ or $xy + yz + zx - 1 = 0$.
Step 4: Conclusion
Hence, the correct relation is $xy+yz+zx-1=0$.
Final Answer:(B)