Đáp án:
$3)
+) \cot \alpha=\dfrac{4}{3}\\
+) \cos\alpha=-\dfrac{4}{5}\\
+)\sin\alpha=-\dfrac{3}{5}\\$
Giải thích các bước giải:
$3)
+) \cot \alpha=\dfrac{1}{\tan \alpha}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\\
+) 1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}\\
\Rightarrow \cos^2\alpha=\dfrac{1}{1+\tan^2\alpha}=\dfrac{1}{1+\left ( \dfrac{3}{4} \right )^2}=\dfrac{16}{25}\\
\Rightarrow \cos \alpha=\pm \dfrac{4}{5}$
Do $\pi<\alpha<\dfrac{3\pi}{2}\Rightarrow \cos\alpha<0$
$\Rightarrow \cos\alpha=-\dfrac{4}{5}\\
+) \tan \alpha=\dfrac{\sin\alpha}{\cos\alpha}\\
\Rightarrow \sin\alpha=\tan\alpha.\cos\alpha=\dfrac{3}{4}.\left (-\dfrac{4}{5} \right )=-\dfrac{3}{5}\\$
