Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
13,\\
\frac{{\cos a + \cos 6a + \cos 11a}}{{\sin a + \sin 6a + \sin 11a}}\\
= \frac{{\left( {\cos a + \cos 11a} \right) + \cos 6a}}{{\left( {\sin a + \sin 11a} \right) + \sin 6a}}\\
= \frac{{2.\cos \frac{{a + 11a}}{2}.\cos \frac{{a - 11a}}{2} + \cos 6a}}{{2.\sin \frac{{a + 11a}}{2}.\cos \frac{{a - 11a}}{2} + \sin 6a}}\\
= \frac{{2\cos 6a.\cos \left( { - 5a} \right) + \cos 6a}}{{2.\sin 6a.\cos \left( { - 5a} \right) + \sin 6a}}\\
= \frac{{\cos 6a.\left( {2\cos \left( { - 5a} \right) + 1} \right)}}{{\sin 6a.\left( {2\cos \left( { - 5a} \right) + 1} \right)}}\\
= \frac{{\cos 6a}}{{\sin 6a}} = \cot 6a\\
14,\\
\frac{{{{\sin }^2}x - {{\tan }^2}x}}{{{{\cos }^2}x - {{\cot }^2}x}} = \frac{{{{\sin }^2}x - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}{{{{\cos }^2}x - \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}} = \frac{{{{\sin }^2}x\left( {1 - \frac{1}{{{{\cos }^2}x}}} \right)}}{{{{\cos }^2}x\left( {1 - \frac{1}{{{{\sin }^2}x}}} \right)}}\\
= \frac{{{{\sin }^2}x.\frac{{{{\cos }^2}x - 1}}{{{{\cos }^2}x}}}}{{{{\cos }^2}x.\frac{{{{\sin }^2}x - 1}}{{{{\sin }^2}x}}}} = \frac{{{{\sin }^2}x.\frac{{ - {{\sin }^2}x}}{{{{\cos }^2}x}}}}{{{{\cos }^2}x.\frac{{ - {{\cos }^2}x}}{{{{\sin }^2}x}}}} = \frac{{{{\sin }^6}x}}{{{{\cos }^6}x}} = {\tan ^6}x\\
15,\\
\frac{{\cos a}}{{1 + \sin a}} + \tan a = \frac{{\cos a}}{{1 + \sin a}} + \frac{{\sin a}}{{\cos a}}\\
= \frac{{{{\cos }^2}a + \sin a\left( {1 + \sin a} \right)}}{{\left( {1 + \sin a} \right).\cos a}}\\
= \frac{{{{\cos }^2}a + \sin a + {{\sin }^2}a}}{{\left( {1 + \sin a} \right).\cos a}}\\
= \frac{{1 + \sin a}}{{\left( {1 + \sin a} \right).\cos a}}\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}a + {{\cos }^2}a = 1} \right)\\
= \frac{1}{{\cos a}}
\end{array}\)
