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

3

We know that the minimum value of $cosec^{-1} x$ is $-\frac{\pi}{2}$ which is attained at x = -1

∴ $cosec^{-1} x + cosec^{-1}y + cosec^{-1} z=-\frac{3\pi}{2}$

$⇒ cosec^{-1} x + cosec^{-1}+cosec^{-1}z = \left(-\frac{\pi}{2}\right)+\left(-\frac{\pi}{2}\right)+\left(-\frac{\pi}{2}\right)$

$⇒ cosec^{-1}x=-\frac{\pi}{2}, cosec^{-1}y = -\frac{\pi}{2}, cosec^{-1}z =-\frac{\pi}{2}$

$ ⇒x = -1,  y = -1, z = -1 $

$∴\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{(-1)}{(-1)}+\frac{(-1)}{(-1)}+\frac{(-1)}{(-1)}= 3$