Prove that
`(cos 3A +2 cos 5 A + cos 7A)/( cos A + 2 cos 3 A + cos 5A) = cos 2 A - sin 2 A tan 3 A.`

Answer

To prove the given expression

\[ \frac{\cos 3A + 2 \cos 5A + \cos 7A}{\cos A + 2 \cos 3A + \cos 5A} = \cos 2A - \sin^2 A \tan 3A, \]

we will start with the left-hand side (LHS) and simplify it step by step.

### Step 1: Write down the LHS
We start with the expression:

\[ \text{LHS} = \frac{\cos 3A + 2 \cos 5A + \cos 7A}{\cos A + 2 \cos 3A + \cos 5A}. \]

### Step 2: Rearrange the terms
We can rearrange the terms in both the numerator and the denominator:

\[ \text{LHS} = \frac{(\cos 7A + \cos 3A) + 2 \cos 5A}{(\cos 5A + \cos A) + 2 \cos 3A}. \]

### Step 3: Apply the cosine addition formula
Using the cosine addition formula \( \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \):

#### Numerator:
Let \( C = 7A \) and \( D = 3A \):

\[ \cos 7A + \cos 3A = 2 \cos\left(\frac{7A + 3A}{2}\right) \cos\left(\frac{7A - 3A}{2}\right) = 2 \cos(5A) \cos(2A). \]

So, the numerator becomes:

\[ \text{Numerator} = 2 \cos(5A) \cos(2A) + 2 \cos(5A) = 2 \cos(5A)(\cos(2A) + 1). \]

#### Denominator:
Let \( C = 5A \) and \( D = A \):

\[ \cos 5A + \cos A = 2 \cos\left(\frac{5A + A}{2}\right) \cos\left(\frac{5A - A}{2}\right) = 2 \cos(3A) \cos(2A). \]

So, the denominator becomes:

\[ \text{Denominator} = 2 \cos(3A) \cos(2A) + 2 \cos(3A) = 2 \cos(3A)(\cos(2A) + 1). \]

### Step 4: Simplify the expression
Now, substituting back into the LHS:

\[ \text{LHS} = \frac{2 \cos(5A)(\cos(2A) + 1)}{2 \cos(3A)(\cos(2A) + 1)}. \]

We can cancel \( 2 \) and \( (\cos(2A) + 1) \) (assuming \( \cos(2A) + 1 \neq 0 \)):

\[ \text{LHS} = \frac{\cos(5A)}{\cos(3A)}. \]

### Step 5: Express \( \cos(5A) \) in terms of \( \cos(3A) \)
Using the cosine addition formula again, we can express \( \cos(5A) \):

\[ \cos(5A) = \cos(3A + 2A) = \cos(3A)\cos(2A) - \sin(3A)\sin(2A). \]

Thus,

\[ \text{LHS} = \frac{\cos(3A) \cos(2A) - \sin(3A) \sin(2A)}{\cos(3A)}. \]

### Step 6: Simplify further
This simplifies to:

\[ \text{LHS} = \cos(2A) - \frac{\sin(3A) \sin(2A)}{\cos(3A)}. \]

Using the identity \( \tan(3A) = \frac{\sin(3A)}{\cos(3A)} \), we can rewrite it as:

\[ \text{LHS} = \cos(2A) - \sin(2A) \tan(3A). \]

### Conclusion
Thus, we have shown that:

\[ \text{LHS} = \cos(2A) - \sin^2 A \tan(3A), \]

which is equal to the right-hand side (RHS) of the original equation. Therefore, we conclude that:

\[ \frac{\cos 3A + 2 \cos 5A + \cos 7A}{\cos A + 2 \cos 3A + \cos 5A} = \cos 2A - \sin^2 A \tan 3A. \]

Hence, the statement is proved.