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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the values of x which satisfy the equation \[ sin^{-1}x+sin^{-1}(1-x)=cos^{-1}x.\]

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Toolbox:
  • \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]\)
  • \( cos^{-1}x=sin^{-1}\sqrt{1-x^2}\)
By taking 1-x in the place of y in the formula of \(sin^{-1}x+sin^{-1}y\)
the given equation can be written as \(sin^{-1}x+sin^{-1}(1-x)=\)
\(sin^{-1}\bigg[x\sqrt{1-(1-x)^2}+(1-x)\sqrt{1-x^2}\bigg]=cos^{-1}x\)
\(\Rightarrow\: sin^{-1} \bigg[ x \sqrt{2x-x^2}+(1-x)\sqrt{1-x^2} \bigg] = cos^{-1}x\)
But we know that \(cos^{-1}x=sin^{-1}\sqrt{1-x^2}\)
\(\Rightarrow\:sin^{-1} \bigg[ x \sqrt{2x-x^2}+(1-x)\sqrt{1-x^2} \bigg] = sin^{-1}[\sqrt{1-x^2} ]\)
\( \Rightarrow x\sqrt{2x-x^2}+(1-x)\sqrt{1-x^2}=\sqrt{1-x^2}\)
\( \Rightarrow x\sqrt{2x-x^2}=\sqrt{1-x^2} (1-(1-x))=\sqrt{1-x^2}.x\)
Squaring both the sides we get
\(x^2(2x-x^2)=x^2(1-x^2)\)
\(\Rightarrow\:x^2(2x-x^2-1+x^2)=0\)
\(x=0\:or\:x=\frac{1}{2}\)
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 20, 2013 by rvidyagovindarajan_1
 

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