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If \( cos^{-1}x+cos^{-1}y+cos^{-1}z=\pi\) prove that \( x^2+y^2+z^2+2xyz=1\)

1 Answer

  • \( cos^{-1}x+cos^{-1}y=cos^{-1} [ xy-\sqrt{1-x^2} \sqrt{1-y^2} ]\)
  • \( cos^{-1}(-x)=\pi -cos^{-1}(x)\)
It is given that \(cos^{-1}x+cos^{-1}y+cos^{-1}z=\pi\)
\(\Rightarrow\: cos^{-1}x+cos^{-1}y=\pi-cos^{-1}z\)
Using the above formula of \(cos^{-1}x+cos^{-1}y\) we get
\( \Rightarrow cos^{-1} [ xy-\sqrt{1-x^2} \sqrt{1-y^2} ] =cos^{-1}(-z)\)
Taking cos on both the sides we get
\( \Rightarrow xy-\sqrt{1-x^2} \sqrt{1-y^2}=cos(cos^{-1} (-z))=-z\)
\( \Rightarrow (xy+z)=\sqrt{1-x^2}.\sqrt{1-y^2}\)
Squaring on both the sides we get
Rearranging the terms we get
\( x^2+y^2+z^2+2xyz=1\)

Hence proved.

answered Mar 1, 2013 by thanvigandhi_1
edited Mar 18, 2013 by rvidyagovindarajan_1

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