Question

Encontre valores de outras funções trigonométricas dadas
$tan \: \alpha = - \frac{1}{3} , \: 90° < \alpha < 180°$
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Answer 1

90°<α<180°→ 3º quadrante

$\boxed{1+{\tan}^{2}(\alpha)={\sec}^{2}(\alpha)}$

$1+{(-\frac{1}{3})}^{2}={\sec}^{2}(\alpha)$

${\sec}^{2}(\alpha)=1+\frac{1}{9}$

${\sec}^{2}(\alpha)=\frac{10}{9}$

$\sec(\alpha)=-\sqrt{\frac{10}{9}}$

$\boxed{\sec(\alpha)=-\frac{\sqrt{10}}{3}}$

$\cos(\alpha)=\frac{1}{\sec(\alpha)}$

$\boxed{\cos(\alpha)=-\frac{3\sqrt{10}}{10}}$

$\boxed{\sin(\alpha)=\tan(\alpha). \cos(\alpha)}$

$\sin(\alpha)=-\frac{1}{3}.-\frac{3\sqrt{10}}{10}$

$\boxed{\sin(\alpha)=\frac{\sqrt{10}}{10}}$

$\boxed{\csc(\alpha)=\frac{1}{\sin(\alpha)}}$

$\boxed{\csc(\alpha)=\sqrt{10}}$

$\boxed{\cot(\alpha)=\frac{1}{\tan(\alpha)}}$

$\boxed{\cot(\alpha)=-3}$

Answer 2

Olá, tudo bem?

$tan \: \alpha = - \frac{1}{3} , \: 90° < \alpha < 180°$

$tag \: \alpha = - \frac{1}{3} \\ \frac{sen \: \alpha }{cos \: \alpha } = - \frac{1}{3} \\ sen {}^{2} \alpha \: = \frac{1}{9} \: . \: \frac{9}{10} \\ sen {}^{2} \alpha = \frac{1}{10} \\ sen \: \alpha = \frac{ \sqrt{10} }{10}$

$sen {}^{2} \: \alpha + cos {}^{2} \: \alpha = 1 \\ ( \frac{ \sqrt{10} }{10} ) {}^{2} \: + \: cos {}^{2} \: \alpha = 1 \\ \frac{1}{10} \: + \: cos {}^{2} \: \alpha = 1 \\ cos {}^{2} \: \alpha = 1 - \frac{1}{10} \\ cos {}^{2} \: \alpha = \frac{9}{10} \\ cos \: \alpha = - \frac{3 \sqrt{10} }{10}$

$cot \: \alpha = \frac{1}{tan \: \alpha } = \frac{1}{ - \frac{1}{3} } = - \frac{3}{1} = - 3$

Espero ter ajudado!