c,
$\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{4}{3}$
* Chứng minh công thức $\dfrac{1}{\cos^2\alpha}=1+\tan^2\alpha$:
VT=$\dfrac{1}{\cos^2\alpha}$
$=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}$
$=\dfrac{\sin^2\alpha}{\cos^2\alpha}+\dfrac{\cos^2\alpha}{\cos^2\alpha}$
$=\tan^2\alpha+1$
$=$ VP
Từ công thức trên, ta có:
$\dfrac{1}{\cos^2\alpha}=1+(\dfrac{3}{4})^2$
$\Leftrightarrow \cos\alpha=0,8$
