## Trigonometry (11th Edition) Clone

$x=2csc^{-1}[(\frac{y+\sqrt 3}{2})]$
$y=-\sqrt 3+2csc\frac{x}{2}$ $y+\sqrt 3=2csc\frac{x}{2}$ Divide by 2 on both sides. $2csc\frac{x}{2}=\frac{y+\sqrt 3}{2}$ $csc\frac{x}{2}=(\frac{y+\sqrt 3}{2})$ Hence, $x=2csc^{-1}[(\frac{y+\sqrt 3}{2})]$