Question:medium

If $3 \sin \alpha = 5 \sin \beta$, then $\tan \left( \frac{\alpha+\beta}{2} \right) + \tan \left( \frac{\alpha-\beta}{2} \right) =$

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Use sum-to-product identities in such questions.
Updated On: May 14, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
We are given a ratio relationship between $\sin \alpha$ and $\sin \beta$. The problem asks for the evaluation of an expression involving tangent functions of half-angles. This specific structure is a classic application of the "Componendo and Dividendo" rule combined with sum-to-product trigonometric identities. (Note: The operator in the problem image appears as a `+` due to fading/blurring, but mathematically it must be a division sign `\div` to evaluate to a constant independent of $\alpha$ and $\beta$. The solution proceeds assuming the standard identity structure which uses division). Step 2: Key Formula or Approach:
1. Componendo and Dividendo rule: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$. 2. Sum-to-product formulas: $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$ $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ Step 3: Detailed Explanation:
Given the relation: $3 \sin \alpha = 5 \sin \beta$. Let's rewrite this as a ratio fraction: \[ \frac{\sin \alpha}{\sin \beta} = \frac{5}{3} \] Now, apply the Componendo and Dividendo rule to both sides: \[ \frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{5 + 3}{5 - 3} \] \[ \frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{8}{2} = 4 \] Next, apply the trigonometric sum-to-product identities to the numerator and denominator on the left side: \[ \frac{2 \sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)}{2 \cos\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)} = 4 \] Cancel the common factor of 2, and group the sine and cosine terms corresponding to the same angles: \[ \left( \frac{\sin\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)} \right) \cdot \left( \frac{\cos\left(\frac{\alpha-\beta}{2}\right)}{\sin\left(\frac{\alpha-\beta}{2}\right)} \right) = 4 \] Recognize the definitions of tangent ($\tan = \sin/\cos$) and cotangent ($\cot = \cos/\sin$): \[ \tan\left(\frac{\alpha+\beta}{2}\right) \cdot \cot\left(\frac{\alpha-\beta}{2}\right) = 4 \] Since $\cot x = \frac{1}{\tan x}$, we can rewrite the expression as a division: \[ \frac{\tan\left(\frac{\alpha+\beta}{2}\right)}{\tan\left(\frac{\alpha-\beta}{2}\right)} = 4 \] This matches the standard expression $\tan\left(\frac{\alpha+\beta}{2}\right) \div \tan\left(\frac{\alpha-\beta}{2}\right)$, whose value is 4. Step 4: Final Answer:
The value is 4.
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