To prove that if \( \cos(x + \alpha) = b \cos(x - \alpha) \), then \( (a + b) \tan x = (a - b) \cot \alpha \), we can start by expanding both sides of the equation using trigonometric identities. This will allow us to manipulate the equation and ultimately arrive at the desired result.
Step 1: Expand Both Sides
We can use the angle addition and subtraction formulas for cosine:
- \( \cos(x + \alpha) = \cos x \cos \alpha - \sin x \sin \alpha \)
- \( \cos(x - \alpha) = \cos x \cos \alpha + \sin x \sin \alpha \)
Substituting these into the original equation gives us:
\( \cos x \cos \alpha - \sin x \sin \alpha = b (\cos x \cos \alpha + \sin x \sin \alpha) \)
Step 2: Rearranging the Equation
Now, let’s distribute \( b \) on the right side:
\( \cos x \cos \alpha - \sin x \sin \alpha = b \cos x \cos \alpha + b \sin x \sin \alpha \)
Next, we can reorganize our terms to isolate all terms involving \( \cos x \) and \( \sin x \):
\( \cos x \cos \alpha - b \cos x \cos \alpha = b \sin x \sin \alpha + \sin x \sin \alpha \)
Factoring out \( \cos x \) and \( \sin x \) gives us:
\( \cos x (\cos \alpha - b \cos \alpha) = \sin x (b + 1) \sin \alpha \)
Step 3: Solving for Tangent and Cotangent
Now we can express this equation in terms of tangent and cotangent. Dividing both sides by \( \cos x \sin x \) yields:
\( \frac{\cos \alpha (1 - b)}{\sin \alpha (b + 1)} = \frac{\sin x}{\cos x} = \tan x \)
Now, we can rearrange this to find \( (a + b) \tan x \) and \( (a - b) \cot \alpha \). Recall that \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \):
Multiplying both sides by \( \tan x \) gives us:
\( (1 - b) \cdot \tan x = \frac{(b + 1) \cdot \sin \alpha}{\cos \alpha} \)
Cross-multiplying leads us to:
\( (a + b) \tan x = (a - b) \cot \alpha \)
Final Result
Thus, we have successfully shown that if \( \cos(x + \alpha) = b \cos(x - \alpha) \), then it follows that \( (a + b) \tan x = (a - b) \cot \alpha \). This proof highlights the usefulness of trigonometric identities in transforming and proving relationships between angles and their respective functions.