`(sinA+sin3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A)` is :

Options

A. `tan 2A`
B. `tan3A`
C. `tan4A`
D. `tan8A`

Answer

To solve the expression \((\sin A + \sin 3A + \sin 5A + \sin 7A) / (\cos A + \cos 3A + \cos 5A + \cos 7A)\), we can follow these steps:

### Step 1: Group the Sine Terms
We can group the sine terms as follows:
\[ \sin A + \sin 7A + \sin 3A + \sin 5A \]
This allows us to use the sine addition formula.

### Step 2: Apply the Sine Addition Formula
Using the formula \(\sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)\):
- For \(\sin A + \sin 7A\):
\[ \sin A + \sin 7A = 2 \sin\left(\frac{A + 7A}{2}\right) \cos\left(\frac{7A - A}{2}\right) = 2 \sin(4A) \cos(3A) \]
- For \(\sin 3A + \sin 5A\):
\[ \sin 3A + \sin 5A = 2 \sin\left(\frac{3A + 5A}{2}\right) \cos\left(\frac{5A - 3A}{2}\right) = 2 \sin(4A) \cos(A) \]

### Step 3: Combine the Sine Terms
Now we can combine the results:
\[ \sin A + \sin 7A + \sin 3A + \sin 5A = 2 \sin(4A) \cos(3A) + 2 \sin(4A) \cos(A) \]
Factoring out \(2 \sin(4A)\):
\[ = 2 \sin(4A) (\cos(3A) + \cos(A)) \]

### Step 4: Group the Cosine Terms
Now, let's group the cosine terms:
\[ \cos A + \cos 3A + \cos 5A + \cos 7A \]
Using the cosine addition formula:
- For \(\cos A + \cos 7A\):
\[ \cos A + \cos 7A = 2 \cos\left(\frac{A + 7A}{2}\right) \cos\left(\frac{7A - A}{2}\right) = 2 \cos(4A) \cos(3A) \]
- For \(\cos 3A + \cos 5A\):
\[ \cos 3A + \cos 5A = 2 \cos\left(\frac{3A + 5A}{2}\right) \cos\left(\frac{5A - 3A}{2}\right) = 2 \cos(4A) \cos(A) \]

### Step 5: Combine the Cosine Terms
Combining these results:
\[ \cos A + \cos 7A + \cos 3A + \cos 5A = 2 \cos(4A) \cos(3A) + 2 \cos(4A) \cos(A) \]
Factoring out \(2 \cos(4A)\):
\[ = 2 \cos(4A) (\cos(3A) + \cos(A)) \]

### Step 6: Form the Final Expression
Now we can substitute back into our original expression:
\[ \frac{2 \sin(4A) (\cos(3A) + \cos(A))}{2 \cos(4A) (\cos(3A) + \cos(A))} \]
The \(2\) and \((\cos(3A) + \cos(A))\) terms cancel out (assuming \(\cos(3A) + \cos(A) \neq 0\)):
\[ = \frac{\sin(4A)}{\cos(4A)} = \tan(4A) \]

### Final Answer
Thus, the expression simplifies to:
\[ \tan(4A) \]