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The value of $\cot^{-1}(-x)$ for all $x\;\in\;R$ in terms of $\cot^{-1}x$ is ____________.

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Toolbox:
  • Principal interval of cot is \((0,\pi)\)
  • \(cot(\pi-\theta)=-cot\theta\)
\(Ans:\:\: \pi - cot^{-1}x\)
Since cot is -ve in the interval \((\frac{\pi}{2},\pi)\), the angle is to be reduced and brought within this interval.
Let \( x=cot\theta,\:\:\Rightarrow\:\theta=cot^{-1}x\)
\(cot^{-1}(-x)=cot^{-1}(-cot\theta)\)
\(=cot^{-1}cot(\pi-\theta)\)
\(=\pi-\theta=\pi-cot^{-1}x\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 10, 2013 by rvidyagovindarajan_1
 
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