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Find the value of x if $cosec^{-1}x=2\cot^{-1}7+\cot^{-1}\frac{3}{4}$

$\begin{array}{1 1} \frac{125}{117} \\ \frac{-125}{117} \\ \frac{-117 }{125} \\ \frac{117}{125} \end{array} $

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Toolbox:
  • \( cot^{-1}x=tan^{-1}\large\frac{1}{x} \)
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} \)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy} \)
  • \( tan^{-1}x=cosec^{-1}\large\frac{\sqrt{1+x^2}}{x} \)
R.H.S.
By substituting \(x=7\:and \:x=\large\frac{4}{3}\)in the formula of \(cot^{-1}x\), R.H.S. becomes
\( 2tan^{-1}\large\frac{1}{7}+tan^{-1}\large\frac{3}{4}\)
 
By taking \(x=\large\frac{1}{7},\large\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{7}}{1-\large\frac{1}{49}}=\large\frac{2}{7}.\large\frac{49}{48}=\large\frac{7}{24}\)
\(\Rightarrow\:2tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{7}{24}\)
\( 2tan^{-1}\large\frac{1}{7}+tan^{-1}\large\frac{3}{4} = tan^{-1}\large\frac{7}{24}+tan^{-1}\large\frac{4}{3} \)
 
By taking\(x=\large\frac{7}{24}\:and\:y=\large\frac{4}{7}\),
\(\large\frac{x+y}{1-xy}=\large\frac{\frac{7}{24}+\large\frac{4}{3}}{1-\large\frac{28}{72}}=\large\frac{117}{72}.\large\frac{72}{44}=\large\frac{117}{44}\)
 
Substituting the values in R.H.S we get
\(tan^{-1}\large\frac{7}{24}+tan^{-1}\large\frac{4}{3} = tan^{-1}\large\frac{\large\frac{7}{24}+\large\frac{4}{3}}{1-\large\frac{28}{72}}=tan^{-1}\large\frac{117}{44} \)
 
By taking \(x=\large\frac{117}{44},\)
\(\large\frac{\sqrt{1+x^2}}{x}=\large\frac{\sqrt{1+\large\frac{13689}{1936}}}{\large\frac{117}{44}}=\sqrt{\large\frac{15624}{1936}}.\large\frac{44}{117}=\large\frac{125}{117}\)
\(\Rightarrow\: tan^{-1}\large\frac{117}{44}= cosec^{-1}\large\frac{125}{117}\)
 
By substituting the value in the above equation we get
\(cosec^{-1}x= cosec^{-1}\frac{125}{117}\)
\(\Rightarrow\: x=\frac{125}{117} \)

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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