Trigonometry (11th Edition) Clone

The value of $y$ here is $$y=\frac{5\pi}{6}$$
$\DeclareMathOperator{\arccot}{arccot}$ $$y=\arccot (-\sqrt 3)$$ First, we see that the domain of inverse cotangent function is $(-\infty,\infty)$. Therefore, in fact when we deal with inverse cotangent function, we do not need to do this checking step. The range of inverse cotangent function is $(0,\pi)$. In other words, $y\in(0,\pi)$. We can rewrite $y=\arccot(-\sqrt 3)$ into $\cot y=-\sqrt3$ We know that $$\cot\frac{\pi}{6}=\sqrt 3$$ which means $$-\cot\frac{\pi}{6}=-\sqrt 3$$ $$\cot(-\frac{\pi}{6})=-\sqrt 3$$ (If you need a proof here, take $\cot(-X)=\frac{\cos(-X)}{\sin(-X)}$. We know that $\cos(-X)=\cos X$ and $\sin(-X)=-\sin X$. That means $\cot(-X)=\frac{\cos X}{-\sin X}=-\cot X$). However, $-\frac{\pi}{6}$ does not belong to the range $(0,\pi)$. So we must take another similar value but stays in quadrant 1 or 2, which belongs to the range. Since $\cot (X+\pi) =\cot x$, $\frac{5\pi}{6}$ is such a value. Therefore, the exact value of $y$ here is $$y=\frac{5\pi}{6}$$