<s>[INST] What are the values of $x$, $y$ and $r$ if $\cot\theta=\frac{5}{6}$ and $\csc\theta <0$? I solved this problem, where $r$ is a radius of a circle. Is my solution right?
What are the values of $x$, $y$ and $r$ if $\cot\theta=\frac{5}{6}$ and $\csc\theta <0$?
Since $\cot\theta$ is positive and $\csc\theta$ is negative, $\theta$ must be on the third quadrant. Therefore, the values of $x$ and $y$ must be both negative. Since $\cot\theta=\frac{\cos\theta}{\sin\theta}$, $\sin\theta=y$ and $\cos\theta=x$, then $x=-5$ and $y=-6$. I can use the Pythagorean Theorem to solve for $r$.
$r^2=x^2+y^2$
$r^2=25+36$
$r^2=61$
$r=\sqrt{61}$
 [/INST] You can't conclude that $x=-5$ and $y=-6$.
You are right that it is in the third quadrant.
$$\cot \theta = \frac{r\cos \theta}{r \sin \theta}=\frac{x}y=\frac{5}{6}.$$
Hence, we have the half-line:
$$6x=5y, x < 0.$$
Any point on the half line satisfies the condition.
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