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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Prove that \( 2sin^{-1}\frac{3}{5}-tan^{-1}\frac{17}{31}=\frac{\pi}{4} \)

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Toolbox:
  • \( sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}\)
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}\: if\: -1 \leq x \leq 1\)
  • \(tan^{-1}x-tan^{-1}y=tan^{-1}\large\frac{x-y}{1+xy} xy > -1\)
By taking \(x=\frac{3}{5},\:\frac{x}{\sqrt{1-x^2}}=\frac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}\)
\(=\large\frac{3}{5}\times\large\frac{5}{4}=\large\frac{3}{4}\)
We know that \(sin^{-1}x=tan^{-1}\frac{x}{\sqrt{1-x^2}}\)
\(\Rightarrow\:\large\:sin^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{3}{4}\)
\( L.H.S= \: 2tan^{-1}\large\frac{3}{4}-tan^{-1}\large\frac{17}{31}\)
We know that \(2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}\)
By taking \(x=\frac{3}{4},\: \frac{2x}{1-x^2}=\frac{2.\frac{3}{4}}{1-\frac{9}{16}}=\frac{3}{2}\times\frac{16}{7}=\frac{24}{7}\)
\(\Rightarrow\:\large\:2tan^{-1}\large\frac{3}{4}=\large\:tan^{-1}\large\frac{24}{7}\)
Substituting the value in L.H.S.
\( \Rightarrow tan^{-1}\large\frac{24}{7}-tan^{-1}\large\frac{17}{31}\)
By taking \(x=\frac{24}{7}\:and\:y=\frac{17}{31}\)
\(\frac{x-y}{1+xy}=\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}.\frac{17}{31}}=1\)
\( \Rightarrow tan^{-1}\large\frac{24}{7}-tan^{-1}\large\frac{17}{31} = tan^{-1}1=\large\frac{\pi}{4}\)
=R.H.S.
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 20, 2013 by rvidyagovindarajan_1
 

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