Cho biết rằng \(0 jwplayer.key="5qMQ1qMprX8KZ79H695ZPnH4X4zDH...

* Hướng dẫn giải

Ta có

\(\begin{array}{l} 1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\\ \Leftrightarrow {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} - 1 = \frac{1}{{{{\left( {\frac{{12}}{{13}}} \right)}^2}}} - 1 = \frac{{25}}{{144}}\\ \Leftrightarrow \tan \alpha = \sqrt {\frac{{25}}{{144}}} = \frac{5}{{12}} \end{array}\)