Question:
If \(3 \sin \alpha=5 \sin \beta\), then \(\tan \left(\frac{\alpha+\beta}{2}\right) \div \tan \left(\frac{\alpha-\beta}{2}\right)=\)
If \(3 \sin \alpha=5 \sin \beta\), then \(\tan \left(\frac{\alpha+\beta}{2}\right) \div \tan \left(\frac{\alpha-\beta}{2}\right)=\)
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Write the ratio of sines and apply componendo and dividendo with sum and difference to product formulas.
Updated On: Nov 13, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Given \(3\sin\alpha=5\sin\beta\), the ratio \(\frac{\sin\alpha}{\sin\beta}=\frac{5}{3}\). We use \(\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\) and the matching difference formula.
Step 2: Apply componendo and dividendo
\(\frac{\sin\alpha+\sin\beta}{\sin\alpha-\sin\beta}=\frac{5+3}{5-3}=4\).
Step 3: Convert to tangents
\(\frac{\sin\alpha+\sin\beta}{\sin\alpha-\sin\beta}=\frac{2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}=\frac{\tan\frac{\alpha+\beta}{2}}{\tan\frac{\alpha-\beta}{2}}\).
Step 4: Result
So \(\tan\frac{\alpha+\beta}{2}\div\tan\frac{\alpha-\beta}{2}=4\), which is option (D).
Final Answer:
The required ratio equals 4. \[ \boxed{\text{(D) }4} \]
Given \(3\sin\alpha=5\sin\beta\), the ratio \(\frac{\sin\alpha}{\sin\beta}=\frac{5}{3}\). We use \(\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\) and the matching difference formula.
Step 2: Apply componendo and dividendo
\(\frac{\sin\alpha+\sin\beta}{\sin\alpha-\sin\beta}=\frac{5+3}{5-3}=4\).
Step 3: Convert to tangents
\(\frac{\sin\alpha+\sin\beta}{\sin\alpha-\sin\beta}=\frac{2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}=\frac{\tan\frac{\alpha+\beta}{2}}{\tan\frac{\alpha-\beta}{2}}\).
Step 4: Result
So \(\tan\frac{\alpha+\beta}{2}\div\tan\frac{\alpha-\beta}{2}=4\), which is option (D).
Final Answer:
The required ratio equals 4. \[ \boxed{\text{(D) }4} \]
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