KCET · Maths · Trigonometric Ratios & Identities
If \(\sqrt{\frac{1+\cos \mathrm{A}}{1-\cos \mathrm{A}}}=\frac{\mathrm{x}}{\mathrm{y}}\), then the value of \(\tan \mathrm{A}\) is equal to
- 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
Given, \(\sqrt{\frac{1+\cos \mathrm{A}}{1-\cos \mathrm{A}}}=\frac{\mathrm{x}}{\mathrm{y}}\)
\(\therefore \sqrt{\frac{2 \cos ^{2} \frac{\mathrm{A}}{2}}{2 \sin ^{2} \frac{\mathrm{A}}{2}}=\frac{\mathrm{x}}{\mathrm{y}}}\)
\(\Rightarrow \quad \tan \frac{\mathrm{A}}{2}=\frac{\mathrm{y}}{\mathrm{x}}\)
Now,
\[
\tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}=\frac{\frac{2 y}{x}}{1-\left(\frac{y}{x}\right)^{2}}
\]
\[
=\frac{2 x y}{x^{2}-y^{2}}
\]
\(\therefore \sqrt{\frac{2 \cos ^{2} \frac{\mathrm{A}}{2}}{2 \sin ^{2} \frac{\mathrm{A}}{2}}=\frac{\mathrm{x}}{\mathrm{y}}}\)
\(\Rightarrow \quad \tan \frac{\mathrm{A}}{2}=\frac{\mathrm{y}}{\mathrm{x}}\)
Now,
\[
\tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}=\frac{\frac{2 y}{x}}{1-\left(\frac{y}{x}\right)^{2}}
\]
\[
=\frac{2 x y}{x^{2}-y^{2}}
\]
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