If `x gt y gt 0`, then find the value of `tan^(-1).(x)/(y) + tan^(-1) [(x + y)/(x -y)]`

Explanation

To solve the problem, we need to find the value of the expression:

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

### Step 1: Use the formula for the sum of inverse tangents

We can use the formula for the sum of two inverse tangent functions:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

provided that \(ab < 1\).

Let \(a = \frac{x}{y}\) and \(b = \frac{x+y}{x-y}\).

### Step 2: Calculate \(ab\)

First, we need to check if \(ab < 1\):

\[ ab = \frac{x}{y} \cdot \frac{x+y}{x-y} = \frac{x(x+y)}{y(x-y)} \]

### Step 3: Calculate \(a + b\)

Now, we calculate \(a + b\):

\[ a + b = \frac{x}{y} + \frac{x+y}{x-y} = \frac{x(x-y) + y(x+y)}{y(x-y)} \]

Simplifying the numerator:

\[ x(x-y) + y(x+y) = x^2 - xy + yx + y^2 = x^2 + y^2 \]

So,

\[ a + b = \frac{x^2 + y^2}{y(x-y)} \]

### Step 4: Calculate \(1 - ab\)

Next, we calculate \(1 - ab\):

\[ 1 - ab = 1 - \frac{x(x+y)}{y(x-y)} = \frac{y(x-y) - x(x+y)}{y(x-y)} \]

Simplifying the numerator:

\[ y(x-y) - x(x+y) = yx - y^2 - x^2 - xy = -x^2 - y^2 \]

Thus,

\[ 1 - ab = \frac{- (x^2 + y^2)}{y(x-y)} \]

### Step 5: Substitute into the formula

Now we can substitute \(a + b\) and \(1 - ab\) into the formula:

\[ \tan^{-1}\left(\frac{a + b}{1 - ab}\right) = \tan^{-1}\left(\frac{\frac{x^2 + y^2}{y(x-y)}}{\frac{-(x^2 + y^2)}{y(x-y)}}\right) \]

This simplifies to:

\[ \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Conclusion

Thus, the value of the expression is:

\[ \tan^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{x+y}{x-y}\right) = -\frac{\pi}{4} \]