Suppose x and y real number such that `tan x tan y=42 and cot x +cot =49` the value of `tan (x+y)` is ____________

Explanation

To solve the problem, we need to find the value of \( \tan(x+y) \) given that \( \tan x \tan y = 42 \) and \( \cot x + \cot y = 49 \).

### Step 1: Convert cotangent to tangent
We know that:
\[ \cot x = \frac{1}{\tan x} \quad \text{and} \quad \cot y = \frac{1}{\tan y} \]
Thus, we can rewrite the equation \( \cot x + \cot y = 49 \) as:
\[ \frac{1}{\tan x} + \frac{1}{\tan y} = 49 \]
Let \( \tan x = a \) and \( \tan y = b \). Then we have:
\[ \frac{1}{a} + \frac{1}{b} = 49 \]
This can be rewritten as:
\[ \frac{a + b}{ab} = 49 \]

### Step 2: Substitute the known product of tangents
We know from the problem that \( ab = \tan x \tan y = 42 \). Substituting this into the equation gives:
\[ \frac{a + b}{42} = 49 \]
Multiplying both sides by 42, we find:
\[ a + b = 49 \times 42 \]
Calculating \( 49 \times 42 \):
\[ 49 \times 42 = 2058 \]
Thus, we have:
\[ a + b = 2058 \]

### Step 3: Use the tangent addition formula
Now we can use the tangent addition formula:
\[ \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
Substituting \( a \) for \( \tan x \) and \( b \) for \( \tan y \):
\[ \tan(x+y) = \frac{a + b}{1 - ab} \]
Substituting the values we found:
\[ \tan(x+y) = \frac{2058}{1 - 42} \]
Calculating the denominator:
\[ 1 - 42 = -41 \]
Thus:
\[ \tan(x+y) = \frac{2058}{-41} \]
Calculating the division:
\[ \tan(x+y) = -50.5 \]

### Final Answer
The value of \( \tan(x+y) \) is:
\[ \boxed{-50.5} \]