a)Ta có:

\(\begin{array}{l}
VP = \frac{1}{{{{\cos }^2}x - {{\sin }^2}x}} + \frac{{2\frac{{{\mathop{\rm sinx}\nolimits} }}{{{\mathop{\rm cosx}\nolimits} }}}}{{1 - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = \frac{1}{{{{\cos }^2}x - {{\sin }^2}x}} + \frac{{2\sin x{\mathop{\rm cosx}\nolimits} }}{{{{\cos }^2}x - {{\sin }^2}x}}\\
 = \frac{{1 + 2\sin x{\mathop{\rm cosx}\nolimits} }}{{\left( {{\mathop{\rm sinx}\nolimits}  + cosx} \right)\left( {\cos x - {\mathop{\rm sinx}\nolimits} } \right)}} = \frac{{{{\left( {\sin  + \cos x} \right)}^2}}}{{\left( {{\mathop{\rm sinx}\nolimits}  + cosx} \right)\left( {\cos x - {\mathop{\rm sinx}\nolimits} } \right)}} = \frac{{\sin  + \cos x}}{{\cos x - {\mathop{\rm sinx}\nolimits} }}\left( 1 \right)\\
Mà \,\,VT = \frac{{\cos 2x}}{{1 - \sin 2x}} = \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x + {{\sin }^2}x - 2sinxcosx}} = \frac{{\left( {\cos x - {\mathop{\rm sinx}\nolimits} } \right)\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)}}{{{{\left( {\cos x - {\mathop{\rm sinx}\nolimits} } \right)}^2}}} = \frac{{\cos x + {\mathop{\rm sinx}\nolimits} }}{{\cos x - {\mathop{\rm sinx}\nolimits} }}\left( 2 \right)
\end{array}\)

Từ (1) và (2) ta được VT = VP =>ĐPCM

b) Ta có:

\(\left\{ \begin{array}{l}
{x^2} - 4x > 5\\
{x^2} - \left( {m - 1} \right)x - m \le 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 5\\
x <  - 1
\end{array} \right.\\
\left( {x + 1} \right)\left( {x - m} \right) \le 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 5\\
x \le m
\end{array} \right. \Leftrightarrow m > 5\\
\left[ \begin{array}{l}
x <  - 1\\
x \ge m
\end{array} \right. \Leftrightarrow m <  - 1
\end{array} \right.\)

Vậy \(\left[ \begin{array}{l}
m <  - 1\\
m > 5
\end{array} \right.\) thì hệ bất phương trình luôn có nghiệm