If sin ^{-1} x+sin ^{-1} y+sin ^{-1} z=frac{3 | vedclass

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(A) We know that the range of $\sin ^{-1} 	heta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since the sum of three such values is $\frac{3 \pi}{2}$,each te

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$3$

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(A) We know that the range of $\sin ^{-1} 	heta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since the sum of three such values is $\frac{3 \pi}{2}$,each term must be equal to its maximum value,$\frac{\pi}{2}$.Thus,$\sin ^{-1} x = \frac{\pi}{2}$,$\sin ^{-1} y = \frac{\pi}{2}$,and $\sin ^{-1} z = \frac{\pi}{2}$.This implies $x = \sin(\frac{\pi}{2}) = 1$,$y = \sin(\frac{\pi}{2}) = 1$,and $z = \sin(\frac{\pi}{2}) = 1$.Substituting these values into the expression,we get $x^{100} + y^{100} + z^{100} = 1^{100} + 1^{100} + 1^{100} = 1 + 1 + 1 = 3$.

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$,then $x^{100}+y^{100}+z^{100}=$

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