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Find the value of the expression $\sin \bigg(2\tan^{-1}{\frac{1}{3}}\bigg)+\cos \big(\tan^{-1}2\sqrt 2\big)$

$\begin{array}{1 1} \frac{15}{14} \\ \frac{-15}{14} \\ \frac{14}{15} \\ \frac{-14}{15} \end{array} $

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Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} \)
  • \( tan^{-1}x=sin^{-1}\large\frac{x}{\sqrt{1+x^2}} \)
  • \( tan^{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}} \)
  •  
By taking x=\(\large\frac{1}{3},\:\large\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{3}}{1-\large\frac{1}{9}}=\large\frac{2}{3}.\large\frac{9}{8}=\large\frac{3}{4}\)
 
Substituting in the formula of \(2tan^{-1}x \) we get
\(2tan^{-1}\large\frac{1}{3}=tan^{-1}\large\frac{3}{4}\)
 
By taking x=\(\large\frac{3}{4},\:\large\frac{x}{\sqrt{1+x^2}}=\large\frac{\large\frac{3}{4}}{\sqrt{1+\large\frac{9}{16}}}=\large\frac{3}{4}.\large\frac{4}{5}=\large\frac{3}{5}\)
 
Substituting in the above formula of \(tan^{-1}x\) we get
\(tan^{-1}\large\frac{3}{4}=sin^{-1}\large\frac{3}{5}\)
 
Similarly by taking\( x=2\sqrt 2,\:\large\frac{1}{\sqrt{1+x^2}}=\large\frac{1}{\sqrt{1+8}}=\large\frac{1}{3}\)
 
Substituting in the above formula of \(tan^{-1}x\) we get
\(tan^{-1} 2\sqrt 2=cos^{-1}\large\frac{1}{3}\)
 
Substituting tthe values in the given expression we get
\( sin \bigg( tan^{-1}\large\frac{3}{4} \bigg) +cos (tan^{-1} 2\sqrt 2) \)
\( = sin \bigg( sin^{-1}\large\frac{3}{5} \bigg)+cos \bigg(cos^{-1}\large\frac{1}{3} \bigg) \)
\( = \large\frac{3}{5}+\frac{1}{3}=\large\frac{14}{15} \)

 

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