A) \[\pm \frac{5}{3}\]
B) \[\pm \frac{\sqrt{5}}{3}\]
C) \[\pm \frac{5}{\sqrt{3}}\]
D) None of these
Correct Answer: B
Solution :
Given that \[\tan \{{{\cos }^{-1}}(x)\}=\sin \left( {{\cot }^{-1}}\frac{1}{2} \right)\] Let \[{{\cot }^{-1}}\frac{1}{2}=\varphi \Rightarrow \frac{1}{2}=\cot \varphi \] \[\Rightarrow \sin \varphi =\frac{1}{\sqrt{1+{{\cot }^{2}}\varphi }}=\frac{2}{\sqrt{5}}\] Let \[{{\cos }^{-1}}x=\theta \Rightarrow \sec \theta =\frac{1}{x}\Rightarrow \tan \theta =\sqrt{{{\sec }^{2}}\theta -1}\] \[\Rightarrow \tan \theta =\sqrt{\frac{1}{{{x}^{2}}}-1}\Rightarrow \tan \theta =\frac{\sqrt{1-{{x}^{2}}}}{x}\] So, \[\tan \{{{\cos }^{-1}}(x)\}=\sin \left( {{\cot }^{-1}}\frac{1}{2} \right)\] \[\Rightarrow \tan \left( {{\tan }^{-1}}\frac{\sqrt{1-{{x}^{2}}}}{x} \right)=\sin \left( {{\sin }^{-1}}\frac{2}{\sqrt{5}} \right)\] \[\Rightarrow \frac{\sqrt{1-{{x}^{2}}}}{x}=\frac{2}{\sqrt{5}}\Rightarrow \sqrt{(1-{{x}^{2}})5}=2x\] Squaring both sides, we get \[x=\pm \frac{\sqrt{5}}{3}\].
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