A) \[{{\cot }^{-1}}x\]
B) \[{{\cot }^{-1}}\frac{1}{x}\]
C) \[{{\tan }^{-1}}x\]
D) None of these
Correct Answer: C
Solution :
The given expression \[=2{{\tan }^{-1}}[\cos ec{{\tan }^{-1}}x-\tan {{\cot }^{-1}}x]\] \[=2{{\tan }^{-1}}\left[ \cos ec\left\{ \cos e{{c}^{-1}}\frac{\sqrt{1+{{x}^{2}}}}{x} \right\} \right.\] \[\left. -\tan \left\{ {{\tan }^{-1}}\left( \frac{1}{x} \right) \right\} \right]\] \[=2{{\tan }^{-1}}\left[ \frac{\sqrt{1+{{x}^{2}}}}{x}-\frac{1}{x} \right]=2{{\tan }^{-1}}\left[ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right]\] \[=2{{\tan }^{-1}}\left[ \frac{\sec \theta -1}{\tan \theta } \right]\] [putting\[x=\tan \theta \]] \[={{\tan }^{-1}}\left[ \frac{1-\cos \theta }{\sin \theta } \right]\] \[=2{{\tan }^{-1}}\left[ \frac{2{{\sin }^{2}}\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}} \right]\] \[=2{{\tan }^{-1}}\tan \frac{\theta }{2}=2\cdot \frac{\theta }{2}=\theta ={{\tan }^{-1}}x\]
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