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