The result `tan^(-1)x-tan^(-1)y = tan^(-1)((x-y)/(1+xy))` is true when the value of xy is `"………."`

Explanation

To determine the necessary condition for the equation \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \) to hold true, we need to analyze the conditions under which the formula is valid.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The formula \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \) is a known identity in trigonometry. For this identity to be valid, certain conditions must be satisfied.

2. **Identifying the Denominator**:
In the right-hand side of the equation, we have a fraction \( \frac{x - y}{1 + xy} \). The denominator \( 1 + xy \) must not be zero to avoid undefined behavior in the equation.

3. **Setting the Condition**:
For the denominator \( 1 + xy \) to be non-zero, we need:
\[ 1 + xy \neq 0 \]
This implies:
\[ xy \neq -1 \]

4. **Ensuring Positivity**:
Additionally, for the tangent function to be defined and for the formula to hold true, we typically require that the product \( xy \) is such that it does not lead to any contradictions in the tangent function's range.

5. **Final Condition**:
Therefore, the necessary condition for the equation to hold is:
\[ xy > -1 \]

### Conclusion:
The result \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \) is true when the value of \( xy \) is greater than \(-1\).

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