This question has confused me for a long time. I would appreciate a brief proof, though a longer proof for deeper comprehension is welcome.
I am given that $$ \lim_{\theta \to 0} k \theta \ \mathrm{cosec}\ \theta = \lim_{\theta \to 0} \theta \ \mathrm{cosec} \ k \theta $$
How do I prove that $k = \pm 1$?
EDIT: There was an error in my textbook. Corrected the right hand side limit.
Answer for the original question
In order $\csc(k\theta)$ makes sense, we have to assume $k\ne0$ throughout.
Since $$ \lim_{\theta\to0}\theta\csc\theta= \lim_{\theta\to0}\frac{\theta}{\sin\theta}=1 $$ the left-hand side has limit $k$.
On the other hand, $$ \lim_{\theta\to0^+}\csc(k\theta)= \begin{cases} \infty & \text{if $k>0$}\\ -\infty & \text{if $k<0$} \end{cases} $$ For the limit $\theta\to0^-$, exchange $\infty$ with $-\infty$ in the above.
So there's no value of $k$ where these limits are the same.
Answer for the edited question
If the right-hand side is $\lim_{\theta\to0}\theta\csc(k\theta)$, we can say $$ \lim_{\theta\to0}\theta\csc(k\theta)= \lim_{\theta\to0}\frac{\theta}{\sin(k\theta)}= \frac{1}{k}\lim_{\theta\to0}\frac{k\theta}{\sin(k\theta)}= \frac{1}{k} $$ Now, equality of the limits gives $$ k=\frac{1}{k} $$ so $k=\pm1$.