The value of ` (sin 7 A + sin 3 A)/(cos 7 A + cos 3 A)` is equal to

Options

A. `tan 5 A`
B. `cot 5 A`
C. `sec 5A - tan 5A`
D. `sec 5A + cot 5A`

Answer

To solve the expression \((\sin 7A + \sin 3A) / (\cos 7A + \cos 3A)\), we can use the sum-to-product identities for sine and cosine.

### Step-by-Step Solution:

1. **Apply the Sum-to-Product Identity for Sine:**
The sum-to-product identity for sine states:
\[ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \]
Here, let \(C = 7A\) and \(D = 3A\). Thus:
\[ \sin 7A + \sin 3A = 2 \sin\left(\frac{7A + 3A}{2}\right) \cos\left(\frac{7A - 3A}{2}\right) = 2 \sin(5A) \cos(2A) \]

2. **Apply the Sum-to-Product Identity for Cosine:**
The sum-to-product identity for cosine states:
\[ \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \]
Using the same values for \(C\) and \(D\):
\[ \cos 7A + \cos 3A = 2 \cos\left(\frac{7A + 3A}{2}\right) \cos\left(\frac{7A - 3A}{2}\right) = 2 \cos(5A) \cos(2A) \]

3. **Substituting Back into the Expression:**
Now substituting these results back into the original expression:
\[ \frac{\sin 7A + \sin 3A}{\cos 7A + \cos 3A} = \frac{2 \sin(5A) \cos(2A)}{2 \cos(5A) \cos(2A)} \]

4. **Canceling Common Terms:**
The \(2\) and \(\cos(2A)\) terms cancel out (assuming \(\cos(2A) \neq 0\)):
\[ = \frac{\sin(5A)}{\cos(5A)} \]

5. **Final Result:**
This simplifies to:
\[ = \tan(5A) \]

Thus, the value of \(\frac{\sin 7A + \sin 3A}{\cos 7A + \cos 3A}\) is \(\tan(5A)\).