\(\sin \alpha = \frac{5}{{13}}\)
A \(\cos \alpha = \frac{{12}}{{13}}\,\,;\,\,\tan \alpha = \frac{5}{{12}}\,\,;\,\,\cot \alpha = \frac{{12}}{5}\)
B \(\cos \alpha = \pm \frac{{12}}{{13}}\,\,;\,\,\tan \alpha = \pm \frac{5}{{12}}\,\,;\,\,\cot \alpha = \pm \frac{{12}}{5}\)
C \(\cos \alpha = \pm \frac{{12}}{{13}}\,\,;\,\,\tan \alpha = \pm \frac{{12}}{5}\,\,;\,\,\cot \alpha = \pm \frac{5}{{12}}\)
D \(\cos \alpha = \frac{{12}}{{13}}\,\,;\,\,\tan \alpha = \frac{{12}}{5}\,\,;\,\,\cot \alpha = \frac{5}{{12}}\)
Hướng dẫn Chọn đáp án là: B
Phương pháp giải:
Sử dụng công thức lượng giác: \(\left\{ \begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\tan \alpha .\cot \alpha = 1\\1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\end{array} \right.\)
Lời giải chi tiết:
\(\sin \alpha = \frac{5}{{13}}\)
Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1 \Leftrightarrow {\left( {\frac{5}{{13}}} \right)^2} + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\cos ^2}\alpha = 1 – \frac{{25}}{{169}} = \frac{{144}}{{169}}\)\( \Rightarrow \cos \alpha = \pm \frac{{12}}{{13}}\)
Lại có: \({\tan ^2}\alpha + 1 = \frac{1}{{{{\cos }^2}\alpha }}\) \( \Leftrightarrow {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} – 1 = \frac{{169}}{{144}} – 1 = \frac{{25}}{{144}}\) \( \Rightarrow \tan \alpha = \pm \frac{5}{{12}}\)
\( \Rightarrow \cot \alpha = \frac{1}{{\tan \alpha }} = \pm \frac{{12}}{5}\)
Chọn B.