`(cot A+tanB)/(cotB+tanA)` is :

Options

A. `tan A cot B`
B. `cot A tan B`
C. 1
D. none of these

Answer

To solve the expression \((\cot A + \tan B) / (\cot B + \tan A)\), we will follow these steps:

### Step 1: Rewrite cotangent and tangent in terms of sine and cosine
We know that:
\[ \cot A = \frac{\cos A}{\sin A} \quad \text{and} \quad \tan B = \frac{\sin B}{\cos B} \]
Thus, we can rewrite the expression:
\[ \cot A + \tan B = \frac{\cos A}{\sin A} + \frac{\sin B}{\cos B} \]

### Step 2: Find a common denominator for the numerator
The common denominator for \(\sin A\) and \(\cos B\) is \(\sin A \cos B\). Therefore, we can express the numerator as:
\[ \frac{\cos A \cos B + \sin A \sin B}{\sin A \cos B} \]

### Step 3: Rewrite cotangent and tangent for the denominator
Similarly, we rewrite \(\cot B\) and \(\tan A\):
\[ \cot B + \tan A = \frac{\cos B}{\sin B} + \frac{\sin A}{\cos A} \]
Again, we find a common denominator for the denominator:
\[ \frac{\cos B \cos A + \sin B \sin A}{\sin B \cos A} \]

### Step 4: Substitute the expressions into the original fraction
Now we can substitute the simplified forms of the numerator and denominator back into the original expression:
\[ \frac{\frac{\cos A \cos B + \sin A \sin B}{\sin A \cos B}}{\frac{\cos B \cos A + \sin B \sin A}{\sin B \cos A}} \]

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{\cos A \cos B + \sin A \sin B}{\cos B \cos A + \sin B \sin A} \cdot \frac{\sin B \cos A}{\sin A \cos B} \]
Notice that \(\cos A \cos B + \sin A \sin B\) is equal to \(\cos(A - B)\) using the cosine addition formula. Thus, we have:
\[ \frac{\sin B \cos A}{\sin A \cos B} \cdot \cos(A - B) \]

### Step 6: Final simplification
The final expression can be simplified further, but we can also observe that:
\[ \frac{\sin B}{\sin A} \cdot \frac{\cos A}{\cos B} \cdot \cos(A - B) \]
This indicates that the original expression simplifies to a function of angles A and B.

### Final Result
Thus, the value of \((\cot A + \tan B) / (\cot B + \tan A)\) simplifies to:
\[ \frac{\sin B}{\sin A} \cdot \frac{\cos A}{\cos B} \cdot \cos(A - B) \]