AP EAMCET · Maths · Trigonometric Ratios & Identities
If \(\cos \alpha+\cos \beta=a, \sin \alpha+\sin \beta=b\) and \(\alpha-\beta=2 \theta\), then \(\frac{\cos 3 \theta}{\cos \theta}=\)
- A \(a^2+b^2-2\)
- B \(a^2+b^2-3\)
- C \(3-a^2-b^2\)
- D \(\frac{a^2+b^2}{4}\)
Answer & Solution
Correct Answer
(B) \(a^2+b^2-3\)
Step-by-step Solution
Detailed explanation
\((\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2 = a^2+b^2\) \(2+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta) = a^2+b^2\) \(2+2\cos(\alpha-\beta) = a^2+b^2\) \(2+2\cos(2\theta) = a^2+b^2\) \(2(2\cos^2\theta) = a^2+b^2\) \(4\cos^2\theta = a^2+b^2\)…
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