1: \(-\pi<\alpha<-\frac{\pi}{2}\)
=>\(cos\alpha<0;\tan\alpha>0;\cot\alpha>0\)
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(cos^2\alpha=1-\left(-\frac34\right)^2=1-\frac{9}{16}=\frac{7}{16}\)
mà \(cos\alpha<0\)
nên \(cos\alpha=-\frac{\sqrt7}{4}\)
\(\tan\alpha=\frac{\sin\alpha}{cos\alpha}=\frac{-3}{4}:\frac{-\sqrt7}{4}=\frac{3}{\sqrt7}\)
\(\cot\alpha=\frac{1}{\tan\alpha}=1:\frac{3}{\sqrt7}=\frac{\sqrt7}{3}\)
2: \(\frac32\pi<\alpha<2\pi\)
=>\(\sin\alpha<0;cos\alpha>0;\tan\alpha<0;\cot\alpha<0\)
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(\sin^2\alpha=1-\left(\frac15\right)^2=1-\frac{1}{25}=\frac{24}{25}\)
mà \(\sin\alpha<0\)
nên \(\sin\alpha=-\sqrt{\frac{24}{25}}=-\frac{2\sqrt6}{5}\)
\(\tan\alpha=\frac{\sin\alpha}{cos\alpha}=\frac{-2\sqrt6}{5}:\frac15=-2\sqrt6\)
\(\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{-2\sqrt6}=\frac{-1}{2\sqrt6}=-\frac{\sqrt6}{12}\)
3: \(\frac{\pi}{2}<\alpha<\pi\)
=>\(\sin\alpha>0;cos\alpha<0;\tan\alpha<0;\cot\alpha<0\)
Ta có: \(1+\tan^2\alpha=\frac{1}{cos^2\alpha}\)
=>\(\frac{1}{cos^2\alpha}=1+\left(-6\right)^2=1+36=37\)
=>\(cos^2\alpha=\frac{1}{37}\)
mà \(cos\alpha<0\)
nên \(cos\alpha=-\frac{1}{\sqrt{37}}\)
\(\sin^2\alpha+cos^2\alpha=1\)
=>\(\sin^2\alpha=1-\frac{1}{37}=\frac{36}{37}\)
mà \(\sin\alpha>0\)
nên \(\sin\alpha=\sqrt{\frac{36}{37}}=\frac{6\sqrt{37}}{37}\)
\(\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{-6}=-\frac16\)
4: \(0<\alpha<\frac{\pi}{2}\)
=>\(\tan\alpha>0;cos\alpha>0;\sin\alpha>0\)
Ta có: \(1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\)
=>\(\frac{1}{\sin^2\alpha}=1+7^2=50\)
=>\(\sin^2\alpha=\frac{1}{50}\)
=>\(\sin\alpha=\sqrt{\frac{1}{50}}=\sqrt{\frac{2}{100}}=\frac{\sqrt2}{10}\)
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(cos^2\alpha=1-\frac{1}{50}=\frac{49}{50}\)
mà \(cos\alpha>0\)
nên \(cos\alpha=\sqrt{\frac{49}{50}}=\frac{7}{5\sqrt2}=\frac{7\sqrt2}{10}\)
\(\tan\alpha=\frac{1}{\cot\alpha}=\frac17\)
