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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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2
We know that $\left|\sin ^{-1} x\right| \leq \frac{\pi}{2}$
Hence, from the given relation we observe that each of $\sin ^{-1} x, \sin ^{-1} y$ and $\sin ^{-1} x$ will be $\frac{\pi}{2}$
so that $x=y=x=\sin \frac{\pi}{2}=1$
$\begin{aligned} \therefore \quad & x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} \\ &=(1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}} \\ &=1+1+1-\frac{1}{1 \times 1 \times 1} \\ &=3-1=2 \end{aligned}$
Hence, from the given relation we observe that each of $\sin ^{-1} x, \sin ^{-1} y$ and $\sin ^{-1} x$ will be $\frac{\pi}{2}$
so that $x=y=x=\sin \frac{\pi}{2}=1$
$\begin{aligned} \therefore \quad & x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} \\ &=(1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}} \\ &=1+1+1-\frac{1}{1 \times 1 \times 1} \\ &=3-1=2 \end{aligned}$
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