Question

If $\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(z) = \frac{\pi}{2}$, then $xy + yz + zx =$

• A. 0
• B. 1
• C. -1
• D. $xyz$

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: The triple inverse-tangent formula.
There is a known identity: \[ \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\!\left(\frac{x + y + z - xyz}{1 - (xy + yz + zx)}\right) \]

Step 2: Use the given sum.
The three inverse tangents add to $\frac{\pi}{2}$.

Step 3: What $\frac{\pi}{2}$ means here.
$\tan$ of $\frac{\pi}{2}$ is infinite. A fraction becomes infinite only when its bottom is zero.

Step 4: Set the denominator to zero.
\[ 1 - (xy + yz + zx) = 0 \]

Step 5: Solve.
\[ xy + yz + zx = 1 \]

Step 6: Conclusion.
\[ \boxed{ xy + yz + zx = 1 } \]