A) \[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\]
B) \[\frac{2xy}{{{x}^{2}}+{{y}^{2}}}\]
C) \[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\]
D) \[\frac{2xy}{{{y}^{2}}-{{x}^{2}}}\]
Correct Answer: C
Solution :
Given, \[\sqrt{\frac{a+\operatorname{cosA}}{1-\cos A}}=\frac{x}{y}\] \[\therefore \] \[\sqrt{\frac{2{{\cos }^{2}}\frac{A}{2}}{2si{{n}^{2}}\frac{A}{2}}}=\frac{x}{y}\] \[\Rightarrow \] \[\tan \frac{A}{2}=\frac{y}{x}\] Now, \[\tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}}=\frac{\frac{2y}{x}}{1-{{\left( \frac{y}{x} \right)}^{2}}}\] \[=\frac{2xy}{{{x}^{2}}-{{y}^{2}}}\]
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