Biểu thức Q = 1 + s i n ( 4 a ) − c o s ( 4 a ) 1 + s i n ( 4 a ) + c o s ( 4 a ) bằng biểu thức nào sau đây:

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Biểu thức\[{\rm{Q = }}\frac{{{\rm{1 + sin}}\left( {{\rm{4a}}} \right) - {\rm{cos}}\left( {{\rm{4a}}} \right)}}{{{\rm{1 + sin}}\left( {{\rm{4a}}} \right){\rm{ + cos}}\left( {{\rm{4a}}} \right)}}\]bằng biểu thức nào sau đây:

\[{\rm{A = sin}}\left( {{\rm{2a}}} \right)\]

\[{\rm{B = cos}}\left( {{\rm{2a}}} \right)\]

\[{\rm{C = tan}}\left( {{\rm{2a}}} \right)\]

\[{\rm{D = cot}}\left( {{\rm{2a}}} \right)\]

Giải thích

\[{\rm{Q = }}\frac{{{\rm{1 + sin}}\left( {{\rm{4a}}} \right) - {\rm{cos}}\left( {{\rm{4a}}} \right)}}{{{\rm{1 + sin}}\left( {{\rm{4a}}} \right){\rm{ + cos}}\left( {{\rm{4a}}} \right)}}{\rm{ = }}\frac{{{\rm{sin}}\left( {{\rm{4a}}} \right){\rm{ + }}\left[ {{\rm{1}} - {\rm{cos}}\left( {{\rm{4a}}} \right)} \right]}}{{{\rm{sin}}\left( {{\rm{4a}}} \right){\rm{ + }}\left[ {{\rm{1 + cos}}\left( {{\rm{4a}}} \right)} \right]}}\]

\[{\rm{ = }}\frac{{{\rm{2sin}}\left( {{\rm{2a}}} \right){\rm{cos}}\left( {{\rm{2a}}} \right){\rm{ + 2si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{2a}}} \right)}}{{{\rm{2sin}}\left( {{\rm{2a}}} \right){\rm{cos}}\left( {{\rm{2a}}} \right){\rm{ + 2co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{2a}}} \right)}}{\rm{ = }}\frac{{{\rm{2sin}}\left( {{\rm{2a}}} \right)\left[ {{\rm{cos}}\left( {{\rm{2a}}} \right){\rm{ + sin}}\left( {{\rm{2a}}} \right)} \right]}}{{{\rm{2cos}}\left( {{\rm{2a}}} \right)\left[ {{\rm{sin}}\left( {{\rm{2a}}} \right){\rm{ + cos}}\left( {{\rm{2a}}} \right)} \right]}}\]

\[{\rm{ = }}\frac{{{\rm{2sin}}\left( {{\rm{2a}}} \right)}}{{{\rm{2cos}}\left( {{\rm{2a}}} \right)}}{\rm{ = tan}}\left( {{\rm{2a}}} \right)\]

Đáp án cần chọn là: C