Respuesta :
[tex]\sin\theta= \frac{4}{5}
\\
\\\sin^2\theta+\cos^2\theta=1
\\ \cos\theta=\pm \sqrt{1-\sin^2\theta} =\pm \sqrt{1-( \frac{4}{5})^2 } =\pm \sqrt{1- \frac{16}{25} } =\pm \sqrt{\frac{9}{25} }=\pm \frac{3}{5}} [/tex]
[tex]\theta \in II \Rightarrow \cos\theta\ \textless \ 0 \Rightarrow \cos\theta=-\frac{3}{5}[/tex]
[tex]\cot\theta= \frac{\cos\theta}{\sin\theta}= \frac{-\frac{3}{5}}{\frac{4}{5}} =- \frac{3}{4} [/tex]
[tex]\theta \in II \Rightarrow \cos\theta\ \textless \ 0 \Rightarrow \cos\theta=-\frac{3}{5}[/tex]
[tex]\cot\theta= \frac{\cos\theta}{\sin\theta}= \frac{-\frac{3}{5}}{\frac{4}{5}} =- \frac{3}{4} [/tex]
