Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
0^\circ < \alpha < 90^\circ \Rightarrow 0^\circ < 2\alpha < 180^\circ \Rightarrow \sin 2\alpha > 0\\
\sin 2\alpha > 0 \Rightarrow \sin 2\alpha = \sqrt {1 - {{\cos }^2}2\alpha } = \sqrt {1 - {{\left( { - \frac{7}{{25}}} \right)}^2}} = \frac{{24}}{{25}}\\
\tan 2\alpha = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} = - \frac{{24}}{7}\\
\cot 2\alpha = \frac{{\cos 2\alpha }}{{\sin 2\alpha }} = - \frac{7}{{24}}\\
b,\\
0^\circ < \alpha < 90^\circ \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha > 0
\end{array} \right.\\
\cos 2\alpha = 2{\cos ^2}\alpha - 1 \Leftrightarrow \frac{{ - 7}}{{25}} = 2{\cos ^2}\alpha - 1 \Rightarrow {\cos ^2}\alpha = \frac{9}{{25}} \Rightarrow \cos \alpha = \frac{3}{5}\,\,\,\,\,\left( {\cos \alpha > 0} \right)\\
\sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } = \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \frac{4}{5}\\
\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{4}{3}\\
\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{3}{4}
\end{array}\)
