A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
\[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\frac{\pi }{2}\] Put\[{{\sin }^{-1}}x=\alpha \], \[{{\sin }^{-1}}y=\beta ,\] \[{{\sin }^{-1}}z=\gamma \] \[\therefore \] \[\alpha +\beta +\gamma =\frac{\pi }{2}\], (Given) or \[\alpha +\beta =\frac{\pi }{2}-\gamma \] or \[\cos (\alpha +\beta )=\cos \left( \frac{\pi }{2}-\gamma \right)\] \[\cos \alpha \cos \beta -\sin \alpha \sin \beta =\sin \gamma \] ?..(i) and, we have \[\sin \alpha =x\] Þ \[\cos \alpha =\sqrt{1-{{x}^{2}}}\] Similarly, \[\cos \beta =\sqrt{1-{{y}^{2}}}\] \[\therefore \] From equation (i), we get \[\sqrt{1-{{x}^{2}}}.\sqrt{1-{{y}^{2}}}=xy+z\] Squaring both sides, we have\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1\].
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