Question

If $ \sin(\alpha + \beta) = 5 \sin(\alpha - \beta) $, then $ \frac{\sin 2\beta}{5 - \cos 2\beta} $ is:

• A. \( \tan(\alpha + \beta) \)
• B. \( \cot(\alpha + \beta) \)
• C. \( \cot(\alpha - \beta) \)
• D. \( \tan(\alpha - \beta) \)

Hint:

Use trigonometric identities and formulas to simplify the given equation and match the required form.

Answer 1

The Correct Option is D

We are given: \[ \sin(\alpha + \beta) = 5 \sin(\alpha - \beta) \] Using the addition and subtraction formulas for sine: \[ \sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \] \[ \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta \] Substitute these into the given equation: \[ \sin\alpha \cos\beta + \cos\alpha \sin\beta = 5(\sin\alpha \cos\beta - \cos\alpha \sin\beta) \] Simplify: \[ \sin\alpha \cos\beta + \cos\alpha \sin\beta = 5 \sin\alpha \cos\beta - 5 \cos\alpha \sin\beta \] Rearrange the equation: \[ (1 - 5) \sin\alpha \cos\beta = (-5 - 1) \cos\alpha \sin\beta \] \[ -4 \sin\alpha \cos\beta = -6 \cos\alpha \sin\beta \] Thus, \[ \frac{\sin\alpha}{\cos\alpha} = \frac{3}{2} \] So, we get \( \tan(\alpha) = \frac{3}{2} \). This can be used to simplify the expression \( \frac{\sin 2\beta}{5 - \cos 2\beta} \) to \( \tan(\alpha - \beta) \).
Thus, the answer is \( \boxed{\tan(\alpha - \beta)} \).