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If $ \sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi,$ prove that $x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

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  • \(cosx-cosy=2sin\large\frac{x+y}{2}.cos\large\frac{y-x}{2}\)
  • \(1-sin^2x=cos^2x\)
  • \(sin(2\pi-\theta)=-sin\theta\)
  • \(sin2A=2sinAcosA\)
  • \(cos(\pi-\theta)=-cos\theta\)
Let \( sin^{-1}x=A, \: sin^{-1}y=B, \: sin^{-1}z=C\)
By substituting the value of x,y,z in the given eqn.
\(\Rightarrow\:given\: A+B+C=\pi\)
Substituting the value of x y and z in L.H.S. we get
\(x \sqrt{1-x^2}+y\sqrt{1-y^2}+2\sqrt{1-z^2}=\)
\(sinA \sqrt{1-sin^2A}+SinB\sqrt{1-sin^2B}+\)
Multiply and divide by 2
Using the formula of sin2A=2sinA.cosA
By taking x=2A,y=2B in the above formula of sinx+siny, we get
By taking \(\theta=C\) in the above formula of \(sin(\pi-\theta)\)
By writing cosC=cos(\(\pi-(A+B))=-cos(A+B)\)
By taking x=A-B, and y=(A+B) in the above formula of cox+cosy we get


answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1

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