Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
31,\\
T = \tan x + \frac{{\cos x}}{{1 + \sin x}} = \frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{1 + \sin x}}\\
= \frac{{\sin x\left( {1 + \sin x} \right) + {{\cos }^2}x}}{{\cos x\left( {1 + \sin x} \right)}}\\
= \frac{{\sin x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{\cos x\left( {1 + \sin x} \right)}}\\
= \frac{{\sin x + 1}}{{\cos x\left( {1 + \sin x} \right)}}\\
= \frac{1}{{\cos x}}\\
32,\\
\frac{\pi }{2} < \alpha < \pi \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha < 0
\end{array} \right.\\
\tan \alpha = \frac{{ - 15}}{7} \Leftrightarrow \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{{15}}{7} \Leftrightarrow \sin \alpha = - \frac{{15}}{7}\cos \alpha \\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( { - \frac{{15}}{7}\cos \alpha } \right)^2} + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = \frac{{49}}{{274}}\\
\Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha = \frac{{225}}{{274}}\\
\sin \alpha > 0 \Rightarrow \sin \alpha = \frac{{15}}{{\sqrt {274} }}\\
33,\\
{\left( {\frac{{\sin \alpha + \tan \alpha }}{{\cos \alpha + 1}}} \right)^2} + 1\\
= {\left( {\frac{{\sin \alpha + \frac{{\sin \alpha }}{{\cos \alpha }}}}{{\cos \alpha + 1}}} \right)^2} + 1\\
= {\left( {\frac{{\frac{{\sin \alpha .\cos \alpha + \sin \alpha }}{{\cos \alpha }}}}{{\cos \alpha + 1}}} \right)^2} + 1\\
= {\left( {\frac{{\sin \alpha \left( {\cos \alpha + 1} \right)}}{{\cos \alpha \left( {\cos \alpha + 1} \right)}}} \right)^2} + 1\\
= {\left( {\frac{{\sin \alpha }}{{\cos \alpha }}} \right)^2} + 1\\
= \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{1}{{{{\cos }^2}\alpha }}
\end{array}\)
