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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Prove that : $ \cos^{-1}\large \frac{4}{5}+\tan^{-1}\large\frac{3}{5}$$=\tan^{-1}\large\frac{27}{11} $

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Toolbox:
  • \( cos^{-1}x=tan^{-1}\large\frac{\sqrt{1-x^2}}{x}\)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1\)
L.H.S
Express\( cos^{-1}\large\frac{4}{5}\)  in terms of  \( tan^{-1} \), using the formula above
\( cos^{-1}x=tan^{-1}\large\frac{\sqrt{1-x^2}}{x}\) by tyaking \( x=\large\frac{4}{5}\) .
That is \( cos^{-1}\large\frac{4}{5}=tan^{-1}\bigg(\large\frac{\sqrt{1-\frac{16}{25}}}{\large\frac{4}{5}}\bigg)\)
\(=tan^{-1}\bigg(\large\frac{\sqrt{\large\frac{9}{25}}}{\large\frac{4}{5}}\bigg)\)
\(=tan^{-1}\bigg(\large\frac{3}{5}\times\frac{5}{4}\bigg)\)=\(tan^{-1}\frac{3}{4}\)
Substituting the value of \( cos^{-1}x\) in L.H.S. we get
 
\( tan^{-1}\large\frac{3}{4}+tan^{-1}\large\frac{3}{5}\)
Using the above formula of \(tan^{-1}x+tan^{-1}y\)  by taking \(x=\large\frac{3}{4}\:\:and\:\:y=\large\frac{3}{5}\) we get
\(= tan^{-1} \bigg( \large\frac{\frac{3}{4}+\large\frac{3}{5}}{1-\large\frac{9}{20}} \bigg)\)
\( = tan^{-1}\large\frac{27}{11}\)

 

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

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