Question

If \( \sin(\theta+\phi) = n \sin(\theta-\phi), \; n \neq 1 \), then the value of \( \frac{\tan\theta}{\tan\phi} \) is

• A. \( \frac{n}{n-1} \)
• B. \( \frac{n+1}{n-1} \)
• C. \( \frac{n}{1-n} \)
• D. \( \frac{n-1}{n+1} \)
• E. \( \frac{1+n}{1-n} \)

Hint:

Always expand trigonometric expressions before simplifying ratios.

Answer 1

The Correct Option is B

Concept: Use identities: \[ \sin(A+B)=\sin A\cos B + \cos A\sin B \] \[ \sin(A-B)=\sin A\cos B - \cos A\sin B \]

Step 1: Expand both sides
\[ \sin\theta\cos\phi + \cos\theta\sin\phi = n(\sin\theta\cos\phi - \cos\theta\sin\phi) \]

Step 2: Bring like terms together
\[ \sin\theta\cos\phi - n\sin\theta\cos\phi = -n\cos\theta\sin\phi - \cos\theta\sin\phi \]

Step 3: Factor
\[ (1-n)\sin\theta\cos\phi = -(n+1)\cos\theta\sin\phi \]

Step 4: Rearrangement
\[ \frac{\sin\theta}{\cos\theta} \cdot \frac{\cos\phi}{\sin\phi} = \frac{n+1}{n-1} \]

Step 5: Convert to tangent
\[ \frac{\tan\theta}{\tan\phi} = \frac{n+1}{n-1} \] \[ \boxed{\frac{n+1}{n-1}} \]