A) \[\sqrt{2}\]
B) \[\sqrt{3}\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: B
Solution :
\[\cos x=\frac{2\cos y-1}{2-\cos y}\] \[\Rightarrow \] \[\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}=\frac{2\left( \frac{1-{{\tan }^{2}}\frac{y}{2}}{1+{{\tan }^{2}}\frac{y}{2}} \right)-1}{2-\left( \frac{1-{{\tan }^{2}}\frac{y}{2}}{1+{{\tan }^{2}}\frac{y}{2}} \right)}\] \[\Rightarrow \] \[\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}=\frac{2-2{{\tan }^{2}}\frac{y}{2}-1-{{\tan }^{2}}\frac{y}{2}}{2+2{{\tan }^{2}}\frac{y}{2}-1+{{\tan }^{2}}\frac{y}{2}}\] \[\Rightarrow \] \[\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}=\frac{1-3{{\tan }^{2}}\frac{y}{2}}{1+3{{\tan }^{2}}\frac{y}{2}}\] \[\Rightarrow \] \[1+3{{\tan }^{2}}\frac{y}{2}-{{\tan }^{2}}\frac{x}{2}-3{{\tan }^{2}}\frac{x}{2}{{\tan }^{2}}\frac{y}{2}\] \[=1-3{{\tan }^{2}}\frac{y}{2}+{{\tan }^{2}}\frac{x}{2}-3{{\tan }^{2}}\frac{x}{2}{{\tan }^{2}}\frac{y}{2}\] \[\Rightarrow \] \[6{{\tan }^{2}}\frac{y}{2}=2{{\tan }^{2}}\frac{x}{2}\] \[\Rightarrow \] \[\tan \frac{x}{2}\cdot \cot \frac{y}{2}=\sqrt{3}\]
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