Answer:
Solution given;
Sec θ=-[tex]\frac{13}{12}[/tex]
cotθ< 0,
It lies in second quadrant.
where sin and cosec is positive.
Now
[tex] \frac{1}{cosθ}=-\frac{13}{12}[/tex]
cosθ=[tex]\frac{12}{13}[/tex]
[tex]\frac{b}{h}[/tex]=[tex]\frac{12}{13}[/tex]
b=12
h=13
By using Pythagoras law
p=[tex] \sqrt{13²-12²}=5 [/tex]
Now
exact values of tan θ=[tex]\frac{p}{b}[/tex]=[tex]\frac{5}{12}[/tex]
since it lies in II quadrant
tan θ=-[tex]\frac{5}{12}[/tex]
and
sinθ=[tex]\frac{p}{h}[/tex]=[tex]\frac{5}{13}[/tex]
since it lies in II quadrant
sin θ=[tex]\frac{5}{13}[/tex]
