If `xy =1 +a^(2) , ` show that ` tan ^(-1) ""(1)/(a+x) +tan^-1"" 1/(a+y) = tan ^(-1) ""(1)/(a) , ( x+ y + 2a) ne 0 `

Answer

To solve the problem, we need to show that:

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

given that \(xy = 1 + a^2\).

### Step 1: Write down the left-hand side (LHS)

We start with the left-hand side:

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

### Step 2: Use the formula for the sum of arctangents

Using the formula for the sum of two arctangents:

\[ \tan^{-1}(u) + \tan^{-1}(v) = \tan^{-1}\left(\frac{u + v}{1 - uv}\right) \]

where \(u = \frac{1}{a+x}\) and \(v = \frac{1}{a+y}\), we can rewrite the LHS:

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

### Step 3: Simplify the numerator

Now, let's simplify the numerator:

\[ \frac{1}{a+x} + \frac{1}{a+y} = \frac{(a+y) + (a+x)}{(a+x)(a+y)} = \frac{2a + x + y}{(a+x)(a+y)} \]

### Step 4: Simplify the denominator

Now, simplify the denominator:

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

### Step 5: Substitute \(xy\)

From the given condition \(xy = 1 + a^2\), we substitute this into the denominator:

\[ a^2 + ax + ay + (1 + a^2) - 1 = 2a^2 + ax + ay \]

Thus, the denominator becomes:

\[ \frac{2a^2 + ax + ay}{(a+x)(a+y)} \]

### Step 6: Combine the results

Now, substituting back into the LHS:

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

### Step 7: Factor out \(a\) from the denominator

We can factor out \(a\) from the denominator:

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

### Step 8: Simplify the expression

This simplifies to:

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

### Conclusion

Thus, we have shown that:

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

Hence, we conclude that:

\[ \text{LHS} = \text{RHS} \]

### Final Statement

Therefore, the statement is proved.

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