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Solve the following equation $\cos\big(\tan^{-1}x\big)=\sin\bigg(\cot^{-1}{\frac{3}{4}}\bigg)$

$\begin{array}{1 1} \frac{4}{3} \\ \pm \frac{3}{4} \\ \frac{4}{5} \\ \frac{3}{5} \end{array} $

Can you answer this question?
 
 

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Toolbox:
  • \( tan^{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}}\)
  • \( cot^{-1}x=sin^{-1}\large\frac{1}{\sqrt{1+x^2}}\)
By taking \(x=\large\frac{3}{4},\:\large\frac{1}{\sqrt{1+x^2}}=\large\frac{1}{\sqrt{1+\large\frac{9}{16}}}=\large\frac{4}{5}\)
\(\Rightarrow\:cot^{-1}\large\frac{3}{4}=sin^{-1}\large\frac{4}{5}\)
 
By using the formula of \(tan^{-1}x\) in L.H.S.
Equation becomes \( cos \bigg[ cos^{-1}\large\frac{1}{\sqrt{1+x^2}} \bigg] = sin \bigg[ sin^{-1}\large\frac{1}{\sqrt{1+\large\frac{9}{16}}} \bigg]\)
\(\Rightarrow\:cos \bigg( cos^{-1}\large\frac{1}{\sqrt{1+x^2}} \bigg) = sin \big( sin^{-1}\large\frac{4}{5}\big)\)
\( \Rightarrow \large\frac{1}{\sqrt{1+x^2}} = \large\frac{4}{5} \)
 
\(\Rightarrow\:1+x^2=\large\frac{25}{16}\)
\(\Rightarrow\:x^2=\large\frac{25}{16}-1=\large\frac{9}{16}\)
\(\Rightarrow\: x = \pm \large\frac{3}{4} \)

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
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