If `tan^(-1)x-tan^(-1)y=tan^(-1)A,` then A=

Options

A. `(x-y)/(1+xy)`
B. `(x+y)/(1-xy)`
C. `x-y`
D. `x+y`

Answer

To solve the equation \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} A \), we can use the formula for the difference of inverse tangents:

\[ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \]

Given that \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} A \), we can equate:

\[ \tan^{-1} \left( \frac{x - y}{1 + xy} \right) = \tan^{-1} A \]

Since the inverse tangent function is one-to-one, we can set the arguments equal to each other:

\[ \frac{x - y}{1 + xy} = A \]

Thus, we can express \( A \) as:

\[ A = \frac{x - y}{1 + xy} \]

Now, we can rewrite this equation to find a relationship between \( x \), \( y \), and \( A \).

To find \( A \) in terms of \( x \) and \( y \), we can rearrange the equation:

\[ A(1 + xy) = x - y \]

This gives us:

\[ x - y = A + Axy \]

Now, we can analyze the options provided in the question.

1. \( x - y = 1 + xy \)
2. \( x + y = 1 - xy \)
3. \( x - y \)
4. \( x + y \)

From our derived equation, we see that if we set \( A = 1 \), we get:

\[ x - y = 1 + xy \]

Thus, the correct answer is:

\[ A = x - y = 1 + xy \]