If `((cotAcotB-1))/((cotB+cotA))=x` then the value of x is

Options

A. cot(A-B)
B. tan (A+B)
C. cot(A+B)
D. tan(A-B)

Answer

To solve the equation

\[ \frac{\cot A \cot B - 1}{\cot B + \cot A} = x, \]

we will simplify the left-hand side step by step.

### Step 1: Rewrite cotangent in terms of sine and cosine
Recall that \(\cot A = \frac{\cos A}{\sin A}\) and \(\cot B = \frac{\cos B}{\sin B}\). Thus, we can rewrite the expression:

\[ \cot A \cot B = \frac{\cos A \cos B}{\sin A \sin B}. \]

Substituting this into the equation gives:

\[ \frac{\frac{\cos A \cos B}{\sin A \sin B} - 1}{\frac{\cos B}{\sin B} + \frac{\cos A}{\sin A}}. \]

### Step 2: Simplify the numerator
The numerator becomes:

\[ \frac{\cos A \cos B - \sin A \sin B}{\sin A \sin B}. \]

Using the cosine addition formula, \(\cos(A + B) = \cos A \cos B - \sin A \sin B\), we can rewrite the numerator as:

\[ \frac{\cos(A + B)}{\sin A \sin B}. \]

### Step 3: Simplify the denominator
The denominator simplifies to:

\[ \frac{\cos B \sin A + \cos A \sin B}{\sin A \sin B} = \frac{\sin(A + B)}{\sin A \sin B}. \]

### Step 4: Combine the fractions
Now, substituting back into our equation gives:

\[ \frac{\frac{\cos(A + B)}{\sin A \sin B}}{\frac{\sin(A + B)}{\sin A \sin B}}. \]

The \(\sin A \sin B\) cancels out:

\[ \frac{\cos(A + B)}{\sin(A + B)} = \cot(A + B). \]

### Step 5: Final result
Thus, we have:

\[ x = \cot(A + B). \]

### Conclusion
The value of \(x\) is:

\[ \boxed{\cot(A + B)}. \]

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