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Solve: $\sin\lfloor2\cos^{-1}\{\cot(2\tan^{-1}x)\}\rfloor=0$

$\begin{array}{1 1} \pi \\ -1 \\ 0 \\ -1 \pm \sqrt 2 \end{array} $

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Toolbox:
  • put \( tan^{-1}x=\theta\)
  • \( \Rightarrow x = tan\theta\)
  • \( tan2\theta = \frac{2tan\theta}{1-tan^2\theta}\)
  • \( cot2\theta=\frac{1}{tan2\theta} \)
put \( tan^{-1}x=\theta\: \Rightarrow x = tan\theta\)
\( cot(2tan^{-1}x) =cot2\theta=\frac{1}{tan2\theta}=\frac{1-tan^2\theta}{2tan\theta}\)
\(=\frac{1-x^2}{2x}\)
Substitute the value in the given eqn.
\( sin \bigg[ 2cos^{-1}\{ cot(2tan^{-1}x) \} \bigg] = 0\)
\( \Rightarrow sin\bigg(2cos^{-1} cot 2\theta\bigg)=0\)
\(\Rightarrow\:2cos^{-1}cot2\theta=sin^{-1}0=0\)
\(\Rightarrow\:cos^{-1}cot2\theta=0\)
\( \Rightarrow cot 2\theta=cos0=1\)
Substituting the value of cot2\(\theta\)
\( \Rightarrow \frac{1-x^2}{2x}=1\)
\( \Rightarrow 1-x^2=2x\)
\(\Rightarrow\:x^2+2x-1=0\)
\(\Rightarrow\:-1\pm\sqrt2\)
answered Feb 19, 2013 by thanvigandhi_1
edited Mar 4 by meena.p
 

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