Answer:
Solution given;
Sec θ=-
![(13)/(12)](https://img.qammunity.org/2022/formulas/mathematics/college/7a8sypefpkt2cj61o3c9dzkto01uf40mpk.png)
cotθ< 0,
It lies in second quadrant.
where sin and cosec is positive.
Now
![(1)/(cosθ)=-(13)/(12)](https://img.qammunity.org/2022/formulas/mathematics/college/cr4dhv42c20136e0mqfiqhzynu22ypartz.png)
cosθ=
![(12)/(13)](https://img.qammunity.org/2022/formulas/mathematics/high-school/2938254g1769su1xr6v6eo65wnnw9mjgjo.png)
=
![(12)/(13)](https://img.qammunity.org/2022/formulas/mathematics/high-school/2938254g1769su1xr6v6eo65wnnw9mjgjo.png)
b=12
h=13
By using Pythagoras law
p=
![√(13²-12²)=5](https://img.qammunity.org/2022/formulas/mathematics/college/8a46cnjsrdvih8gtgkr4epknqoni14ybo0.png)
Now
exact values of tan θ=
=
![(5)/(12)](https://img.qammunity.org/2022/formulas/mathematics/high-school/x5hvvrx3xtdj6vb1drrvkhhiubleg5lk.png)
since it lies in II quadrant
tan θ=-
![(5)/(12)](https://img.qammunity.org/2022/formulas/mathematics/high-school/x5hvvrx3xtdj6vb1drrvkhhiubleg5lk.png)
and
sinθ=
=
![(5)/(13)](https://img.qammunity.org/2022/formulas/mathematics/high-school/jr7bvqs9a4x3hhq51wj92ukdz67lxwj7mn.png)
since it lies in II quadrant
sin θ=
![(5)/(13)](https://img.qammunity.org/2022/formulas/mathematics/high-school/jr7bvqs9a4x3hhq51wj92ukdz67lxwj7mn.png)