e: Ta có: \(\frac{\pi}{2}\le\alpha\le\pi\)
=>\(cos\alpha<0;\tan\alpha<0;\cot\alpha<0\)
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(cos^2\alpha=1-\left(\frac{12}{13}\right)^2=1-\frac{144}{169}=\frac{25}{169}\)
mà \(cos\alpha<0\)
nên \(cos\alpha=-\frac{5}{13}\)
\(\tan\alpha=\frac{\sin\alpha}{cos\alpha}=\frac{12}{13}:\frac{-5}{13}=-\frac{12}{5}\)
\(\cot\alpha=\frac{1}{\tan\alpha}=1:\frac{-12}{5}=-\frac{5}{12}\)
f: \(\tan\alpha\cdot\cot\alpha=1\)
=>\(\cot\alpha=\frac13\)
\(\alpha\in\left(\pi;\frac32\pi\right)\)
=>\(\sin\alpha<0;cos\alpha<0\)
Ta có \(1+\tan^2\alpha=\frac{1}{cos^2\alpha}\)
=>\(\frac{1}{cos^2\alpha}=1+3^2=10\)
=>\(cos^2\alpha=\frac{1}{10}\)
=>\(cos\alpha=-\frac{1}{\sqrt{10}}\)
TA có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(\sin^2\alpha=1-\frac{1}{10}=\frac{9}{10}\)
mà \(\sin\alpha<0\)
nên \(\sin\alpha=-\frac{3}{\sqrt{10}}\)
