Chứng minh \[1 + tanx + ta{n^2}x + ta{n^3}x = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x}}{{{{\cos }^3}x}}\].

\[1 + tanx + ta{n^2}x + ta{n^3}x\]

\[8:\frac{5}{4} \times \frac{5}{4} = 8\]

\[ = \frac{{{{\cos }^3}x + {\mathop{\rm s}\nolimits} {\rm{inx}}.{{\cos }^2}x + {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}.\cos x + {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^3}{\rm{x}}}}{{{{\cos }^3}x}}\]

\[ = \frac{{{{\cos }^2}x.({\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x) + {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}.({\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x)}}{{{{\cos }^3}x}}\]

Vậy \[1 + tanx + ta{n^2}x + ta{n^3}x = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x}}{{{{\cos }^3}x}}\].