sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

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​sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

Saurabh Koranglekar · Answer

Dear student Consider x = y= z = pi/3 And check the equation Lhs != Rhs Kindly check the question Regards

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sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

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​sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

​sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

1 Answers

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sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0

sin^-1 x + sin^-1 y +sin^-1 z = pi. prove that x^2-y^2-z^2+2yz[root(1-x^2)]=0