If A,B,C be the angles of a triangle, then `sum (cot A + cot B)/( tan A + tan B ) =`

Options

A. 1
B. `-1`
C. 0
D. none

Answer

To solve the problem, we need to find the value of the expression:

\[ \sum \frac{\cot A + \cot B}{\tan A + \tan B} \]

where \(A\), \(B\), and \(C\) are the angles of a triangle. We know that the sum of the angles in a triangle is \(180^\circ\), i.e., \(A + B + C = 180^\circ\).

### Step 1: Use the identity for cotangent and tangent
We can express cotangent and tangent in terms of sine and cosine:

\[ \cot A = \frac{\cos A}{\sin A}, \quad \tan A = \frac{\sin A}{\cos A} \]

Thus, we can rewrite the expression:

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

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

\[ \frac{\cos A \sin B + \cos B \sin A}{\sin A \sin B} \]

And the denominator becomes:

\[ \frac{\sin A \cos B + \sin B \cos A}{\cos A \cos B} \]

### Step 3: Combine the fractions
Now we can combine the fractions:

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

### Step 4: Recognize the angle sum identities
The numerator can be recognized as:

\[ \cos A \sin B + \cos B \sin A = \sin(A + B) \]

And the denominator can be simplified using the identity \(A + B + C = 180^\circ\) (thus \(A + B = 180^\circ - C\)):

\[ \sin A \cos B + \sin B \cos A = \sin(A + B) = \sin(180^\circ - C) = \sin C \]

### Step 5: Final expression
Thus, we can rewrite our expression as:

\[ \frac{\sin C \cdot \cos A \cos B}{\sin A \sin B \cdot \sin C} \]

Cancelling \(\sin C\) from the numerator and denominator (assuming \(C \neq 0\)) gives us:

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

### Step 6: Final result
This can be further simplified to:

\[ \frac{1}{\tan A \tan B} \]

Thus, the final answer is:

\[ \sum \frac{\cot A + \cot B}{\tan A + \tan B} = 1 \]