In $\triangle ABC$ we have:

$\sin A= 2018 \sin B\sin C \text { and } \cos A= 2018 \cos B \cos C$ ,

then what is the value of $(\tan A-1)=?$

The answer is 2018.

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sin(A)=2018sin(B)sin(C) & cos(A)=2018cos(B)cos(C); hence tan(A)=tan(B)tan(C). Since A+B+C=π we can say that tan(A)+tan(B)+tan(C)=tan(A)tan(B)tan(C)=tan²(A). Moreover, tan²(A) = tan(A) + sin(B)/cos(B) + sin(C)/cos(C) =tan(A) + sin(B+C)/[cos(B)cos(C)] = tan(A) +sin(A) /[cos(A)/2018] =tan(A)+2018tan(A). Cancel tan(A) from both sides since tan(A)≠0 to obtain, tan(A)=1+2018 and thus tan(A) - 1 = 2018