Answer:
[tex]tan(\theta)=-\frac{7}{6}[/tex]
[tex]sec(\theta)=\frac{\sqrt{85} }{6}[/tex]
[tex]cos(\theta)=\frac{6\sqrt{85} }{85}[/tex]
[tex]sin(\theta)=-\frac{7\sqrt{85}}{85}[/tex]
[tex]cosec(\theta)=-\frac{\sqrt{85}}{7}[/tex]
Step-by-step explanation:
[tex]cot (\theta) = -\frac{6}{7}[/tex]
a) Since,
[tex]tan(\theta) = \frac{1}{cot(\theta)}[/tex]
[tex]tan(\theta) = \frac{1}{-\frac{6}{7} }=-\frac{7}{6}[/tex]
b) Also, according to the Pythagorean identity:
[tex]sec^{2}(\theta)=1+tan^{2}(\theta)[/tex]
Using the value of tan([tex]\theta[/tex]), we get:
[tex]sec^{2}(\theta)=1+(-\frac{7}{6} )^{2}\\\\ sec^{2}(\theta)=\frac{85}{36}\\\\ sec(\theta)=\pm \sqrt{\frac{85}{36} } \\\\ sec(\theta)=\pm \frac{\sqrt{85} }{6}[/tex]
Since, secant is positive in 4th quadrant, we will only consider the positive value. i.e.
[tex]sec(\theta)=\frac{\sqrt{85} }{6}[/tex]
c) Since,
[tex]cos(\theta)=\frac{1}{sec(\theta)}[/tex]
Using the value of secant, we get:
[tex]cos(\theta)=\frac{1}{\frac{\sqrt{85} }{6} } =\frac{6\sqrt{85} }{85}[/tex]
d) According to Pythagorean identity:
[tex]sin^{2}(\theta)=1-cos^{2}(\theta)\\sin(\theta)=\pm \sqrt{1-cos^{2}(\theta)}[/tex]
Since, sine is negative in fourth quadrant, we will consider the negative value. Using the value of cosine, we get:
[tex]sin(\theta)=-\sqrt{1-(\frac{6\sqrt{85} }{85})^{2}}=-\frac{7\sqrt{85}}{85}[/tex]
e) Since,
[tex]cosec(\theta)=\frac{1}{sin(\theta)}[/tex]
Using the value of sine, we get:
[tex]cosec(\theta)=\frac{1}{-\frac{7\sqrt{85} }{85}}=-\frac{\sqrt{85}}{7}[/tex]
