Thẻ: Toán lớp 9

\({\rm{cos}}\alpha = \frac{3}{4}\) A \(\sin \alpha = \pm \frac{4}{5}\,\,;\,\,\tan \alpha = \pm \frac{{16}}{{15}}\,\,;\,\,\cot \alpha = \pm \frac{{15}}{{16}}\) B \(\sin \alpha = \frac{4}{5}\,\,;\,\,\tan \alpha = \frac{{16}}{{15}}\,\,;\,\,\cot …

\(\tan \alpha = \frac{{12}}{{35}}\) 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 …

\(\sin \alpha = \frac{5}{{13}}\) 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 …

\(\sin \alpha = \frac{2}{3}\) A \(\cos \alpha = \pm \frac{{\sqrt 5 }}{3}\,\,;\,\,\,\tan \alpha = \pm \frac{{2\sqrt 5 }}{5}\,\,;\,\,\,\cot \alpha = \pm \frac{{\sqrt 5 }}{2}\) B …

\(\tan \alpha = \frac{4}{3}\) A \(\sin \alpha = \pm \frac{4}{5}\,\,;\,\,\cos \alpha = \pm \frac{3}{5}\,\,;\,\,\cot \alpha = \frac{3}{4}\) B \(\sin \alpha = \frac{4}{5}\,\,;\,\,\cos \alpha = \frac{3}{5}\,\,;\,\,\cot …

\({\rm{sin }}{49^o},{\rm{ cos }}{15^o},{\rm{ sin }}{65^o},{\rm{ cos }}{50^o},{\rm{ }}\cos {\rm{ }}{42^o}\) A \(\sin {49^0} < \sin {65^0} < \cos {15^0} < \cos {50^0} < \cos …

Với góc nhọn \(\alpha \) tùy ý, khẳng định nào sau đây là Sai? A \(\tan \,\alpha = \frac{{\cos \alpha }}{{\sin \alpha }}.\) B \(\tan …

Khẳng định nào dưới đây là đúng? A \(\sin \alpha = \cos \left( {{{90}^0} – \alpha } \right)\) B \(\sin {\alpha ^2} + \cos {\alpha …

\(\cos {\rm{ }}{44^o},{\rm{ sin }}{50^o},{\rm{ sin }}{70^o},{\rm{ cos }}{55^o}\) A \(\cos {44^0} < \sin {50^0} < \sin {70^0} < \cos {55^0}\) B \(\cos{44^0} < \cos …

Chứng minh đẳng thức sau đúng với mọi \({0^0} < \alpha < {90^0}\) \(\frac{{{{\cot }^2}\alpha – {{\cos }^2}\alpha }}{{{{\cot }^2}\alpha }} + \frac{{\sin \alpha .cos\alpha …