Find $\tan a \cdot\tan b$ given equation involving $\sin x$ and $\cos x$ 
Given:
  $$(9\sin a + 44\cos a) \cdot (9\sin b + 44\cos b) = 2017$$
  Find  $\tan a \cdot \tan b$.

I factored the given equation, but I don't know how to proceed. Note that $9^2 + 44^2 = 2017$, that should be helpful in some way. 
 A: $$
(9\sin a + 44\cos a) (9\sin b + 44\cos b) = 2017
$$
Since $9^2+44^2 = 2017,$ you have
$$
\left(\frac 9 {\sqrt{2017}} \right)^2 + \left( \frac{44}{\sqrt{2017}} \right)^2 = 1
$$
and so for some value of $\theta$ you have
$$
\cos\theta = \frac 9 {\sqrt{2017}}, \quad \text{and} \quad \sin\theta = \frac{44}{\sqrt{2017}}.
$$
So
$$
(\cos\theta\sin a+ \sin\theta\cos a) (\cos\theta\sin b + \sin\theta\cos b) = 1
$$
$$
\sin(a+\theta) \sin(b+\theta) = 1
$$
Assuming all numbers are real, that can happen only if both sines are $+1$ or both are $-1.$
If $a+\theta$ and $b+\theta$ are both $\pi/2,$ modulo $2\pi,$ then $a$ and $b$ are the same modulo $2\pi.$
Since $\tan\theta = \dfrac{44}9,$ its complementary angle has tangent $\dfrac 9 {44}.$ Thus $\tan a\tan b = (\tan a)^2 = 9^2/44^2.$
Do something similar if both sines are $-1.$
A: hint: CS inequality gives: LHS $\le 2017$. Can you continue? If you need a bit more details, here it is:  $|9\sin x + 44\cos x| \le \sqrt{9^2+44^2} = \sqrt{2017}$. Put $x = a, b$ and gets the right side of $2017$. Equality occurs allows you to calculate $\tan a\tan b$. Hope you can finish it.
