by \(\tan \alpha = \frac{{12}}{{35}}\)
A \(\cot \alpha = \frac{{35}}{{12}}\,\,;\,\,\cos \alpha = \frac{{35}}{{37}}\,\,;\,\,\sin \alpha = \frac{{12}}{{37}}\)
B \(\cot \alpha = \frac{{35}}{{12}}\,\,;\,\,\sin \alpha = \pm \frac{{35}}{{37}}\,\,;\,\,\cos \alpha = \pm \frac{{12}}{{37}}\)
C \(\cot \alpha = \frac{{35}}{{12}}\,\,;\,\,\cos \alpha = \pm \frac{{35}}{{37}}\,\,;\,\,\sin \alpha = \pm \frac{{12}}{{37}}\)
D \(\cot \alpha = \frac{{35}}{{12}}\,\,;\,\,\sin \alpha = \frac{{35}}{{37}}\,\,;\,\,\cos \alpha = \frac{{12}}{{37}}\)
Hướng dẫn Chọn đáp án là: C
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:
\(\tan \alpha = \frac{{12}}{{35}}\)
Ta có: \(\tan \alpha .\cot \alpha = 1\)\( \Leftrightarrow \cot \alpha = \frac{1}{{\tan \alpha }} = 1:\frac{{12}}{{35}} = \frac{{35}}{{12}}\)
Lại có: \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\)\( \Leftrightarrow 1 + {\left( {\frac{{12}}{{35}}} \right)^2} = \frac{1}{{{{\cos }^2}\alpha }}\)\( \Leftrightarrow \frac{1}{{{{\cos }^2}\alpha }} = \frac{{1369}}{{1225}}\)\( \Rightarrow {\cos ^2}\alpha = \frac{{1225}}{{1369}}\)\( \Rightarrow \cos \alpha = \pm \frac{{35}}{{37}}\)
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\left( {\frac{{35}}{{37}}} \right)^2} + {\sin ^2}\alpha = 1\)\( \Leftrightarrow {\sin ^2}\alpha = 1 – \frac{{1225}}{{1369}} = \frac{{144}}{{1369}}\)\( \Rightarrow \sin \alpha = \pm \frac{{12}}{{37}}\)
Chọn C.
