Đáp án:
$\sin\alpha=\dfrac{\sqrt{26}}{26}\\\cos\alpha=\dfrac{5\sqrt{26}}{26}\\\tan\alpha=\dfrac{1}{5}$
Giải thích các bước giải:
Do $\alpha$ là góc nhọn nên $0<\sin \alpha,\cos \alpha<1;\;\tan \alpha,\cot \alpha>0$
Ta có:
+) $\cot\alpha=5\Rightarrow\tan\alpha=\dfrac{1}{5}$
+) $1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}$
$\Leftrightarrow 1+\dfrac{1}{25}=\dfrac{1}{\cos^2\alpha}$
$\Rightarrow\cos^2\alpha=\dfrac{25}{26}$
$\Rightarrow \cos\alpha=\dfrac{5\sqrt{26}}{26}$
+) $\sin^2\alpha+\cos^2\alpha=1$
$\Leftrightarrow \sin^2\alpha+\dfrac{25}{26}=1$
$\Leftrightarrow \sin^2\alpha=\dfrac{1}{26}$
$\Rightarrow \sin\alpha=\dfrac{\sqrt{26}}{26}$
