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If $3\tan^{-1}x+\cot^{-1}x=\pi$ then $x$ equals

$\begin{array}{1 1} 0 \\ 1 \\ -1 \\ \frac{1}{2}\end{array} $

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  • \( tan^{-1}x+cot^{-1}x=\large\frac{\pi}{2} \)
  • \(tan^{-1}1=\large\frac{\pi}{4}\)
Ans (B) 1
 
We can write the given equation as
\(2tan^{-1}x+tan^{-1}x+cot^{-1}x=\large\frac{\pi}{2} \)
 
\( \Rightarrow\:2tan^{-1}x+\large\frac{\pi}{2}=\pi\)
\( \Rightarrow 2tan^{-1}x=\pi-\large\frac{\pi}{2}=\large\frac{\pi}{2} \)
\( \Rightarrow tan^{-1}x=\large\frac{\pi}{4}\)
 
\(\Rightarrow x = 1\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 

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