COMEDK · Maths · 8. Trigonometric Ratios & Identities
If \(\sin \alpha=k \sin \beta\), then \(\tan \left(\dfrac{\alpha-\beta}{2}\right) \cot \left(\dfrac{\alpha+\beta}{2}\right)\) is equal to
- A \(\dfrac{k+1}{k-1}\)
- B \(\dfrac{k-1}{k+1}\)
- C \(\dfrac{1-k}{1+k}\)
- D \(\dfrac{1+k}{1-k}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{k-1}{k+1}\)
Step-by-step Solution
Detailed explanation
Given \(\sin \alpha = k \sin \beta\), we have \(\dfrac{\sin \alpha}{\sin \beta} = k\). Applying componendo and dividendo: \(\dfrac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \dfrac{k+1}{k-1}\) Using the sum-to-product formulas…
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