`(cos7A+cos5A)/(sin7A-sin5A)=`

Options

A. tanA
B. cotA
C. sinA
D. None of these

Explanation

To solve the expression \((\cos 7A + \cos 5A) / (\sin 7A - \sin 5A)\), we can use trigonometric identities.

### Step-by-Step Solution:

1. **Use the Cosine Sum Identity**:
The sum of cosines can be expressed using the identity:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Here, let \(A = 7A\) and \(B = 5A\):
\[ \cos 7A + \cos 5A = 2 \cos\left(\frac{7A + 5A}{2}\right) \cos\left(\frac{7A - 5A}{2}\right) = 2 \cos(6A) \cos(A) \]

2. **Use the Sine Difference Identity**:
The difference of sines can be expressed using the identity:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Again, let \(A = 7A\) and \(B = 5A\):
\[ \sin 7A - \sin 5A = 2 \cos\left(\frac{7A + 5A}{2}\right) \sin\left(\frac{7A - 5A}{2}\right) = 2 \cos(6A) \sin(A) \]

3. **Substituting Back into the Expression**:
Now substitute the results from steps 1 and 2 back into the original expression:
\[ \frac{\cos 7A + \cos 5A}{\sin 7A - \sin 5A} = \frac{2 \cos(6A) \cos(A)}{2 \cos(6A) \sin(A)} \]

4. **Simplifying the Expression**:
The \(2 \cos(6A)\) terms in the numerator and denominator cancel out (assuming \(\cos(6A) \neq 0\)):
\[ = \frac{\cos(A)}{\sin(A)} \]

5. **Final Result**:
The expression simplifies to:
\[ = \cot(A) \]

Thus, we have shown that:
\[ \frac{\cos 7A + \cos 5A}{\sin 7A - \sin 5A} = \cot(A) \]