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:

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

Given that \(x > y > 0\).

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

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

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

In our case, let:
- \(a = \frac{x}{y}\)
- \(b = \frac{x+y}{x-y}\)

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

First, we need to check the product \(ab\):

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

### Step 3: Check the condition \(ab < 1\)

Since \(x > y > 0\), we can simplify \(ab\):

\[ ab = \frac{x^2 + xy}{xy - y^2} \]

Now, we need to determine if this is less than 1. We will proceed with the calculation assuming \(ab < 1\) holds true.

### Step 4: Apply the sum formula

Now, we can apply the sum formula:

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

### Step 5: Simplify the numerator

Calculating the numerator:

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

### Step 6: Simplify the denominator

Now for the denominator:

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

### Step 7: Combine the results

Now we can combine the results:

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

### Step 8: Final simplification

After simplifying, we will find that:

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

### Step 9: Add \(\pi\)

Since we used the sum formula, we need to add \(\pi\):

\[ \tan^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{x+y}{x-y}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

### Final Answer

Thus, the value of the expression is:

\[ \frac{3\pi}{4} \]