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

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Aicharya · Accepted Answer

Given $\begin{array}{l} \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2} \Rightarrow \sin ^{-1} x=\sin ^{-1} y=\sin ^{-1} z=\frac{\pi}{2} \ 	herefore x=y=z=1 \ 	herefore x^{100}+y^{100}+z^{100}=1+1+1=3 \end{array}$

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

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