Câu1. Ta có\(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(\frac{1}{4}\right)^2\)
\(=\frac{15}{16}\Rightarrow\sin\alpha=\frac{\sqrt{15}}{4}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sqrt{15}}{4}:\frac{1}{4}=\sqrt{15}\)\(=4\sin\alpha\)
Câu2.
Ta có: \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}=5\Leftrightarrow\cos\alpha+\sin\alpha=5\cos\alpha-5\sin\alpha\)
\(\Leftrightarrow4\cos\alpha=6\sin\alpha\Leftrightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{2}{3}\)
\(\Rightarrow\tan\alpha=\frac{2}{3}\)
