Respuesta :
Answer:
[tex]\sin\,\theta =-\frac{\sqrt{21} }{5}[/tex]
[tex]\tan\,\theta =\frac{\sqrt{21} }{2}[/tex]
[tex]\sec\,\theta = \frac{-5}{2}[/tex]
[tex]cosec\,\theta =\frac{-5}{\sqrt{21} }[/tex]
[tex]\cot\,\theta =\frac{2}{\sqrt{21} }[/tex]
Step-by-step explanation:
[tex]\cos\theta =\frac{-2}{5}<0\\sin\theta <0[/tex]
As both [tex]sin\,\theta<0,\,\cos\,\theta <0[/tex],
[tex]\theta[/tex] lies in the third quadrant.
In the third quadrant,
[tex]\sin\theta<0,\,\cos\,\theta <0\,,\,\sec\,\theta<0,\,cosec\theta<0,\,tan\,\theta>0,\,cot\,\theta>0[/tex]
[tex]\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}[/tex]
[tex]\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}[/tex]
[tex]\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}[/tex]
[tex]\ cosec \,\theta = \frac{1}{sin\,\theta }\\\\=\frac{1}{\frac{-\sqrt{21} }{5} }\\\\=\frac{-5}{\sqrt{21} }[/tex]
[tex]\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }[/tex]
