If `(tan^(-1)x)^(3)+(tan^(-1)y)^(3)=1-3tan^(-1)x.tan^(-1)y`. Then which of the following may be true (a)`(x+y)/(tan1)=-2` (b)`(tan^(-1)x)/(1-tan^(-1)y)=2` (c)`(tan^(-1)x)/(1-tan^(-1)y)=2` (d)`(x+y)/(cot1)=1`

Options

A. `(x+y)/(tan1)=-2`
B. `(tan^(-1)x)/(1-tan^(-1)y)=2`
C. `(x+y)/(tan1)=2`
D. `(x+y)/(cot1)=1`

Explanation

To solve the given equation \( (\tan^{-1} x)^3 + (\tan^{-1} y)^3 = 1 - 3 \tan^{-1} x \tan^{-1} y \), we can use the identity for the sum of cubes.

### Step-by-Step Solution:

1. **Identify the Components**:
We have \( a = \tan^{-1} x \) and \( b = \tan^{-1} y \). The equation can be rewritten as:
\[ a^3 + b^3 = 1 - 3ab \]

2. **Use the Sum of Cubes Identity**:
The identity states that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
We can rewrite the left side using this identity:
\[ (a + b)(a^2 - ab + b^2) = 1 - 3ab \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ (a + b)(a^2 - ab + b^2) + 3ab = 1 \]

4. **Using Another Identity**:
We know that:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
Thus, we can substitute \( a^2 - ab + b^2 \) as:
\[ (a + b)^2 - 3ab \]
Therefore, our equation becomes:
\[ (a + b)((a + b)^2 - 3ab) + 3ab = 1 \]

5. **Setting Up the Final Equation**:
Let \( s = a + b \) and \( p = ab \). The equation simplifies to:
\[ s(s^2 - 3p) + 3p = 1 \]
This leads to:
\[ s^3 - 3sp + 3p = 1 \]

6. **Finding \( a \) and \( b \)**:
From the original equation, we can also use the identity:
\[ a + b + c = 0 \]
where \( c = -1 \). Thus:
\[ a + b - 1 = 0 \Rightarrow a + b = 1 \]

7. **Finding Values**:
Since \( a = \tan^{-1} x \) and \( b = \tan^{-1} y \), we have:
\[ \tan^{-1} x + \tan^{-1} y = 1 \]
Taking the tangent of both sides gives:
\[ \frac{x + y}{1 - xy} = \tan(1) \]

8. **Solving for \( x + y \)**:
Rearranging gives:
\[ x + y = \tan(1)(1 - xy) \]

9. **Conclusion**:
If we assume \( x = y = -\tan(1) \), then:
\[ x + y = -2\tan(1) \]
Therefore:
\[ \frac{x + y}{\tan(1)} = -2 \]

### Final Answer:
Thus, the correct option is:
**(a)** \( \frac{x+y}{\tan(1)} = -2 \)