Calculating $\tan A$

Question

I was solving this trigonometric problem and I am stuck,

If, In a $\triangle{ABC}$ $$\sin A=2018\cdot\sin B\cdot\sin C$$

$$\cos A=2018\cdot\cos B\cdot\cos C$$ then, $\tan A$ equals?

1. Not finite
2. $1$
3. $0$
4. $2018$
5. $2019$

My Attempt:

I divided both the equations to get, $$\tan A=\tan B\cdot\tan C$$

Also, $$\tan{(B+C)}=\frac{\tan B+\tan C}{1-\tan B\cdot\tan C}$$

$$\tan{(\pi-A)}=\frac{\tan B+\tan C}{1-\tan A}$$

$$-\tan A=\frac{\tan B+\tan C}{1-\tan A}$$

$$\tan^2 A-\tan A=\tan B+\tan C$$

I do not know how to proceed. Any hints would be helpful.

Answer 1

$$\sin{A}=2018\cdot\sin{B}\cdot\sin{C}\tag1$$ $$\cos{A}=2018\cdot\cos{B}\cdot\cos{C}\tag 2$$ $$\cos(B+C)=\cos B\cos C-\sin B\sin C\tag3$$ $$\cos(B+C)=\cos(\pi-A)=-\cos A\tag4$$ From $(1),(2)$ $$\sin A-\cos A=2018(\sin{B}\cdot\sin{C}-\cos{B}\cdot\cos{C})$$ By $(3),(4)$ $$\sin A-\cos A=-2018\cos(B+C)=2018\cos A$$ $$\iff\tan A-1=2018\implies\tan A=2019$$