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