Question

If \( 3\sin(\alpha-\beta) = 5\cos(\alpha+\beta) \) and \( \alpha+\beta \neq \frac{\pi}{2} \), then \( \frac{\tan(\frac{\pi}{4}-\alpha)}{\tan(\frac{\pi}{4}-\beta)} = \)

• A. 0
• B. -4
• C. \( -\frac{1}{4} \)
• D. \( \frac{1}{2} \)

Hint:

Expressing \( \frac{1-\tan x}{1+\tan x} \) as \( \tan(\frac{\pi}{4}-x) \) is a standard transformation. Recognizing the expansion of the product leads directly to sum/difference formulas.

Answer 1

The Correct Option is C

Step 1: Simplify the Required Expression:

Let \( E = \frac{\tan(\frac{\pi}{4}-\alpha)}{\tan(\frac{\pi}{4}-\beta)} \). Using \( \tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x} = \frac{\cos x - \sin x}{\cos x + \sin x} \): \[ E = \frac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha} \cdot \frac{\cos\beta + \sin\beta}{\cos\beta - \sin\beta} \] Group the terms: \[ E = \frac{(\cos\alpha - \sin\alpha)(\cos\beta + \sin\beta)}{(\cos\alpha + \sin\alpha)(\cos\beta - \sin\beta)} \]
Step 2: Expand Numerator and Denominator:

Numerator \( N = \cos\alpha\cos\beta + \cos\alpha\sin\beta - \sin\alpha\cos\beta - \sin\alpha\sin\beta \) Using compound angle formulas: \[ N = (\cos\alpha\cos\beta - \sin\alpha\sin\beta) - (\sin\alpha\cos\beta - \cos\alpha\sin\beta) \] \[ N = \cos(\alpha+\beta) - \sin(\alpha-\beta) \] Denominator \( D = \cos\alpha\cos\beta - \cos\alpha\sin\beta + \sin\alpha\cos\beta - \sin\alpha\sin\beta \) \[ D = (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + (\sin\alpha\cos\beta - \cos\alpha\sin\beta) \] \[ D = \cos(\alpha+\beta) + \sin(\alpha-\beta) \]
Step 3: Substitute Given Condition:

Given \( 3\sin(\alpha-\beta) = 5\cos(\alpha+\beta) \). Implies \( \sin(\alpha-\beta) = \frac{5}{3}\cos(\alpha+\beta) \). Substitute this into \( E = N/D \): \[ E = \frac{\cos(\alpha+\beta) - \frac{5}{3}\cos(\alpha+\beta)}{\cos(\alpha+\beta) + \frac{5}{3}\cos(\alpha+\beta)} \] \[ E = \frac{1 - 5/3}{1 + 5/3} = \frac{-2/3}{8/3} = \frac{-2}{8} = -\frac{1}{4} \]
Step 4: Final Answer:

The value is \( -1/4 \).