TS EAMCET · Maths · Trigonometric Ratios & Identities
If \(\tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \cos ^2 \alpha}\), then \(\tan (\alpha+\beta) \cdot \cot \alpha=\)
- A \(\frac{-1}{n-1}\)
- B \(n+1\)
- C \(1-n\)
- D \(\frac{1}{n+1}\)
Answer & Solution
Correct Answer
(A) \(\frac{-1}{n-1}\)
Step-by-step Solution
Detailed explanation
\(\tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \cos ^2 \alpha}=\frac{n \tan \alpha}{\sec ^2 \alpha-n}=\frac{n \tan \alpha}{(1-n)+\tan ^2 \alpha}\) \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)…
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