If `tan^(-1)x+tan^(-1)y=(pi)/(4)`, then `cot^(-1)x+cot^(-1)y=`

Options

A. `(pi)/(2)`
B. `(3pi)/(4)`
C. `pi`
D. `(5pi)/(4) `

Answer

To solve the problem, we start with the given equation:

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

### Step 1: Use the identity for the sum of inverse tangents
We know that:

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

This holds true when \( xy < 1 \). Since we have:

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

This implies:

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

### Step 2: Set up the equation
From the tangent addition formula, we can set up the equation:

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

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ x + y = 1 - xy \]

### Step 4: Rearranging the equation
Rearranging the equation, we get:

\[ x + y + xy = 1 \]

### Step 5: Use the identity for cotangent
We know that:

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

### Step 6: Substitute the known value
Substituting the known value of \( \tan^{-1} x + \tan^{-1} y \):

\[ \cot^{-1} x + \cot^{-1} y = \frac{\pi}{2} - \frac{\pi}{4} \]

### Step 7: Simplify the expression
This simplifies to:

\[ \cot^{-1} x + \cot^{-1} y = \frac{\pi}{4} \]

### Final Answer
Thus, we conclude that:

\[ \cot^{-1} x + \cot^{-1} y = \frac{3\pi}{4} \]