`(sin x + sin 3x)/(cos x - cos 3x) = ?`

Options

A. `tan x`
B. `tan 2x`
C. `cot x`
D. `cot 2x`

Answer

To solve the expression \((\sin x + \sin 3x) / (\cos x - \cos 3x)\) and prove that it is equal to \(\cot x\), we can follow these steps:

### Step 1: Apply the identities for \(\sin 3x\) and \(\cos 3x\)
We know the identities:
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
\[ \cos 3x = 4 \cos^3 x - 3 \cos x \]
Substituting these identities into the expression, we get:
\[ \frac{\sin x + (3 \sin x - 4 \sin^3 x)}{\cos x - (4 \cos^3 x - 3 \cos x)} \]

### Step 2: Simplify the numerator and denominator
The numerator simplifies to:
\[ \sin x + 3 \sin x - 4 \sin^3 x = 4 \sin x - 4 \sin^3 x \]
The denominator simplifies to:
\[ \cos x - (4 \cos^3 x - 3 \cos x) = \cos x + 3 \cos x - 4 \cos^3 x = 4 \cos x - 4 \cos^3 x \]
Thus, we can rewrite the expression as:
\[ \frac{4 \sin x - 4 \sin^3 x}{4 \cos x - 4 \cos^3 x} \]

### Step 3: Factor out common terms
Factoring out \(4\) from both the numerator and the denominator gives us:
\[ \frac{4(\sin x - \sin^3 x)}{4(\cos x - \cos^3 x)} \]
We can cancel \(4\) from the numerator and denominator:
\[ \frac{\sin x - \sin^3 x}{\cos x - \cos^3 x} \]

### Step 4: Factor the remaining expressions
We can factor \(\sin x\) and \(\cos x\) out of the numerator and denominator:
\[ \sin x(1 - \sin^2 x) \text{ and } \cos x(1 - \cos^2 x) \]
Using the Pythagorean identity, \(1 - \sin^2 x = \cos^2 x\) and \(1 - \cos^2 x = \sin^2 x\), we rewrite the expression as:
\[ \frac{\sin x \cos^2 x}{\cos x \sin^2 x} \]

### Step 5: Simplify the expression
Now we can simplify:
\[ \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \]
Thus, we have:
\[ \cot x \]

### Conclusion
We have shown that:
\[ \frac{\sin x + \sin 3x}{\cos x - \cos 3x} = \cot x \]
Hence, the expression is proved.

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