If cot (x+y)=`1/sqrt3` and cot (x-y)=`sqrt3`, find the smallest positive value of x and y .

Answer

To solve the problem, we are given two equations involving cotangent:

1. \( \cot(x+y) = \frac{1}{\sqrt{3}} \)
2. \( \cot(x-y) = \sqrt{3} \)

We need to find the smallest positive values of \( x \) and \( y \).

### Step 1: Convert cotangent values to angles
From the first equation, we know that:
\[ \cot(x+y) = \frac{1}{\sqrt{3}} \implies x+y = 60^\circ \]
This is because \( \cot(60^\circ) = \frac{1}{\sqrt{3}} \).

### Step 2: Use the second equation
From the second equation, we have:
\[ \cot(x-y) = \sqrt{3} \implies x-y = 30^\circ \]
This is because \( \cot(30^\circ) = \sqrt{3} \).

### Step 3: Set up the system of equations
Now we have a system of two equations:
1. \( x + y = 60^\circ \) (Equation 1)
2. \( x - y = 30^\circ \) (Equation 2)

### Step 4: Add the equations
Adding both equations:
\[ (x+y) + (x-y) = 60^\circ + 30^\circ \]
This simplifies to:
\[ 2x = 90^\circ \]
Dividing both sides by 2 gives:
\[ x = 45^\circ \]

### Step 5: Substitute to find \( y \)
Now, substitute \( x = 45^\circ \) back into Equation 1:
\[ 45^\circ + y = 60^\circ \]
Subtracting \( 45^\circ \) from both sides gives:
\[ y = 15^\circ \]

### Final Answer
The smallest positive values are:
\[ x = 45^\circ, \quad y = 15^\circ \]