A) \[2\sqrt{3}\]
B) \[\sqrt{3}\]
C) \[\frac{1}{\sqrt{3}}\]
D) \[\sqrt{3}-1\]
Correct Answer: B
Solution :
Given, \[{{\cot }^{-1}}x-{{\cot }^{-1}}(x+2)=\frac{\pi }{12};\] if \[x>0\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{1}{x} \right)-{{\tan }^{-1}}\left( \frac{1}{x+2} \right)=\frac{\pi }{12}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left\{ \frac{\frac{1}{x}-\frac{1}{x+2}}{1+\frac{1}{x(x+2)}} \right\}=\frac{\pi }{12}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left\{ \frac{x+2-x}{{{x}^{2}}+2x+1} \right\}=\frac{\pi }{12}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{2}{{{(x+1)}^{2}}} \right)=\frac{\pi }{12}\] \[\Rightarrow \] \[\frac{2}{{{(x+1)}^{2}}}=\tan \left( \frac{\pi }{12} \right)=\tan {{15}^{o}}\] \[\Rightarrow \] \[\frac{2}{{{(1+x)}^{2}}}=\tan ({{45}^{o}}-{{30}^{o}})\] \[=\frac{\tan {{45}^{o}}-\tan {{30}^{o}}}{1+\tan {{45}^{o}}\,\tan {{30}^{o}}}\] \[\Rightarrow \] \[\frac{2}{{{(1+x)}^{2}}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\] \[\Rightarrow \] \[{{(1+x)}^{2}}=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}\] \[{{(1+x)}^{2}}=\frac{2(3+1+2\sqrt{3})}{3-1}=4+2\sqrt{3}\] \[\Rightarrow \] \[{{(1+x)}^{2}}={{(1+\sqrt{3})}^{2}}\] \[\Rightarrow \] \[(1+x)=1+\sqrt{3}\] \[\therefore \] \[x=\sqrt{3}\]
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