AP EAMCET · Maths · Trigonometric Ratios & Identities
If two acute angles \(A\) and \(B\) are such that \(A \neq B\) and \(\frac{x}{y}=\frac{\cos A}{\cos B}\), then \(\frac{x \tan A-y \tan B}{x+y}=\)
- A \(\tan \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
- B \(\tan \left(\frac{B-A}{2}\right)\)
- C \(\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\)
- D \(\cot \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\tan \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
Step-by-step Solution
Detailed explanation
Given : \(\frac{x}{y}=\frac{\cos A}{\cos B}\) \(\frac{x \tan A-y \tan B}{x+y}=\frac{\frac{x}{y} \tan A-\tan B}{\frac{x}{y}+1}\) \(=\frac{\frac{\cos A}{\cos B} \cdot \tan A-\tan B}{\frac{\cos A}{\cos B}+1}\)…
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