COMEDK · Maths · 8. Trigonometric Ratios & Identities
If \(\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}\), then the value of \(\tan A=\)
- A \(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}\)
- B \(\frac{2 x y}{x^{2}+y^{2}}\)
- C \(\frac{2 x y}{x^{2}-y^{2}}\)
- D \(\frac{2 x y}{y^{2}-x^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 x y}{x^{2}-y^{2}}\)
Step-by-step Solution
Detailed explanation
We have, \(\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}\) \(\Rightarrow \sqrt{\frac{2 \cos ^{2} \frac{A}{2}}{2 \sin ^{2} \frac{A}{2}}}=\frac{x}{y}\) \(\Rightarrow \quad \cot \frac{A}{2}=\frac{x}{y} \Rightarrow \tan \frac{A}{2}=\frac{y}{x}\) Now,…
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