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Question:

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

Solution:

We know

$-\frac{\pi}{2} \leq \sin ^{-1} a \leq \frac{\pi}{2}$, for all $a \in[-1,1]$

So, the minimum value of $\sin ^{-1} a$ is $-\frac{\pi}{2}$.

Now,

$\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=-\frac{3 \pi}{2} \quad$ (Given)

This is possible if

$\sin ^{-1} x=-\frac{\pi}{2}, \sin ^{-1} y=-\frac{\pi}{2}$ and $\sin ^{-1} z=-\frac{\pi}{2}$

$\Rightarrow x=-1, y=-1$ and $z=-1$

$\therefore x y z=(-1) \times(-1) \times(-1)=-1$

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

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