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Solve for x : \[ tan^{-1} \bigg( \frac{2x}{1-x^2} \bigg) +cot^{-1} \bigg( \frac{1-x^2}{2x} \bigg) = \frac{\pi}{3}, x > 0\]

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  • Put \( x = tan\theta \Rightarrow \theta=tan^{-1}x\)
  • \( \large\frac{2tan\theta}{1-tan^2\theta}=tan2\theta\)
  • \( \large\frac{1-tan^2\theta}{2tan\theta}=cot2\theta\)
Put x=\(tan\theta\) in L.H,S,
Then L.H.S.= \(\tan{-1}\large\frac{2tan\theta}{1-tan^2\theta}+cot^{-1}\large\frac{1-tan^2\theta}{2tan\theta}\)
Using the above formula \(\large\frac{2tan\theta}{1-tan^2\theta}=tan2\theta\) and
\(\large\frac{1-tan^2\theta}{2tan\theta}=cot2\theta\) we get 
\(\Rightarrow tan^{-1}tan2\theta+cot^{-1}cot2\theta=\large\frac{\pi}{3}\)
\( \Rightarrow 4\theta=\large\frac{\pi}{3}\)
\( \theta =\large \frac{\pi}{12}=tan^{-1}x\)
\( x = tan\large\frac{\pi}{12}\)


answered Mar 1, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1

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