Phương pháp giải:

Tổng quát:

\(\eqalign{  & {1 \over {\cos kx\cos (k + 1)x}} = {1 \over {\sin \,x}}.{{\sin \,x} \over {\cos kx\cos \left( {k + 1} \right)x}} = {1 \over {\sin \,x}}.{{\sin \,\left[ {\left( {k + 1} \right)x - kx} \right]} \over {\cos kx\cos \left( {k + 1} \right)x}}  \cr   &  = {1 \over {\sin \,x}}.{{\sin \left( {k + 1} \right)x\cos kx - \sin kx\cos \left( {k + 1} \right)x} \over {\cos kx\cos \left( {k + 1} \right)x}} = {1 \over {\sin \,x}}.\left[ {\tan \left( {k + 1} \right)x - \tan kx} \right] \cr} \)

Lời giải chi tiết:

Ta có:

\(\eqalign{  & {1 \over {\cos kx\cos (k + 1)x}} = {1 \over {\sin \,x}}.{{\sin \,x} \over {\cos kx\cos \left( {k + 1} \right)x}} = {1 \over {\sin \,x}}.{{\sin \,\left[ {\left( {k + 1} \right)x - kx} \right]} \over {\cos kx\cos \left( {k + 1} \right)x}}  \cr   &  = {1 \over {\sin \,x}}.{{\sin \left( {k + 1} \right)x\cos kx - \sin kx\cos \left( {k + 1} \right)x} \over {\cos kx\cos \left( {k + 1} \right)x}} = {1 \over {\sin \,x}}.\left[ {\tan \left( {k + 1} \right)x - \tan kx} \right] \cr} \)

\(\eqalign{  & T = {1 \over {\cos x\cos 2x}} + {1 \over {\cos 2x\cos 3x}} + ... + {1 \over {\cos (n - 1)x\cos nx}}  \cr   & \,\,\,\, = {1 \over {\sin \,x}}\left[ {\tan 2x - \tan \,x + \tan 3x - \tan 2x + .... + \tan \,nx - \tan (n - 1)x} \right]  \cr   & \,\,\,\, = {1 \over {\sin \,x}}\left[ { - \tan \,x + \tan \,nx} \right]  \cr   & n = 2018,\,\,x = {\pi  \over 6}  \cr   &  \Rightarrow T = {1 \over {\sin \,{\pi  \over 6}}}\left[ {\tan \left( {2018.{\pi  \over 6}} \right) - \tan {\pi  \over 6}} \right] = {1 \over {\sin \,{\pi  \over 6}}}\left[ {\tan \left( {336\pi  + {\pi  \over 3}} \right) - \tan {\pi  \over 6}} \right] = {1 \over {{1 \over 2}}}\left[ {\sqrt 3  - {1 \over {\sqrt 3 }}} \right] = {4 \over {\sqrt 3 }} \cr} \)

Chọn: A.