Answer:
Using trigonometric ratio:
[tex]\sec \theta = \frac{1}{\cos \theta}[/tex]
[tex]\tan \theta = \frac{\sin \theta}{\cos \theta}[/tex]
From the given statement:
[tex]\cos \theta = -\frac{4}{5}[/tex] and sin < 0
⇒[tex]\theta[/tex] lies in the 3rd quadrant.
then;
[tex]\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}[/tex]
Using trigonometry identities:
[tex]\sin \theta = \pm \sqrt{1-\cos^2 \theta}[/tex]
Substitute the given values we have;
[tex]\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}[/tex]
Since, sin < 0
⇒[tex]\sin \theta = -\frac{3}{5}[/tex]
now, find [tex]\tan \theta[/tex]:
[tex]\tan \theta = \frac{\sin \theta}{\cos \theta}[/tex]
Substitute the given values we have;
[tex]\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}[/tex]
Therefore, the exact value of:
(a)
[tex]\sec \theta =-\frac{5}{4}[/tex]
(b)
[tex]\tan \theta= \frac{3}{4}[/tex]
