An inequality involving sine functions

Let $a>0$ be a fixed number and let $\theta\in(0,\pi/2)$ be also fixed. Is the following inequality true? $$(k^2+a)-\frac{(1+a)\sin (k\theta)}{k\sin\theta}\geq 0 \quad \quad \forall k=1,2,\cdots.$$

Of course, it is an equality for $k=1$. Also, it is true for large $k$. This inequality seems to be obvious (if we had $|\sin(k\theta)|\leq k\sin\theta$), but I failed to prove it rigorously. Note that this question asks whether it is true for all $k\geq1$

Thanks

Thanks for your interest in the question. I thought about it for a while, and the answer is yes. It is proved via induction. Obviously, we can assume $\sin(k\theta)\geq0$; otherwise, the inequality is apparent. Then we just need to show $\sin(k\theta)\leq k\sin\theta$. This is true for $k=1$. Now, suppose it is true for $k$. For $k+1$, we have $$\frac{\sin(k+1)\theta}{(k+1)\sin\theta}=\frac{\sin(k\theta)\cos\theta+ \cos(k\theta)\sin\theta}{k\sin\theta}\frac{k}{k+1}\leq (\cos\theta+\frac{1}{k})\frac{k}{k+1}\leq1,$$ which shows the stated inequality is also true for $k+1$. The desired inequality follows.\qed