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

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(C) We know that the range of $\sin ^{-1} 	heta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given that $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}

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$2$

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(C) We know that the range of $\sin ^{-1} 	heta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given that $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$.Since the maximum value of each term $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $\frac{\pi}{2}$, the sum can be $\frac{3 \pi}{2}$ only if $\sin ^{-1} x = \frac{\pi}{2}$, $\sin ^{-1} y = \frac{\pi}{2}$, 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:$x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} = (1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}}$$= 1+1+1-\frac{1}{1 	imes 1 	imes 1}$$= 3-1 = 2$.

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2},$ then the value of $x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}}$ is equal to

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