If sin^{-1} x + sin^{-1} y + sin^{-1} z = fra | 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 term

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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 attain its maximum value.Therefore,$\sin^{-1} x = \frac{\pi}{2}$,$\sin^{-1} y = \frac{\pi}{2}$,and $\sin^{-1} z = \frac{\pi}{2}$.This implies $x = \sin(\frac{\pi}{2}) = 1$,$y = 1$,and $z = 1$.Substituting these values into the expression: $1^{100} + 1^{100} + 1^{100} - \frac{9}{1^{101} + 1^{101} + 1^{101}}$.$= 1 + 1 + 1 - \frac{9}{1 + 1 + 1} = 3 - \frac{9}{3} = 3 - 3 = 0$.

If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}$,then the value of $x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}}$ is equal to

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