Question

If \(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi\), then the value of \(x+y+z-xyz\) is

• A. 1
• B. 0
• C. -1
• D. 1/2

Hint:

Whenever sum of \(\tan^{-1}\) terms equals \(\pi\), directly use: \(x + y + z = xyz\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use the compound formula for the sum of three inverse tangent functions.
: Key Formula or Approach:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1} \left( \frac{x + y + z - xyz}{1 - (xy + yz + zx)} \right) \]
Step 2: Detailed Explanation:
Given: \[ \tan^{-1} \left( \frac{x + y + z - xyz}{1 - (xy + yz + zx)} \right) = \pi \] Taking tangent on both sides: \[ \frac{x + y + z - xyz}{1 - (xy + yz + zx)} = \tan \pi \] We know that \( \tan \pi = 0 \). \[ \frac{x + y + z - xyz}{1 - (xy + yz + zx)} = 0 \] For a fraction to be zero, its numerator must be zero: \[ x + y + z - xyz = 0 \].
Step 3: Final Answer:
The value is 0.