Find the values of the other five trigonometric functions for angle S.

a) cos S=1/3 and the terminal side of angle S is in Q1.
b) tan S=-10/7 and the terminal side of S is in Q2.
c) sin S=-5/14 and the terminal side of S is in Q3.
d) sec S=13/11 and the terminal side of angle S is in Q4.

To find the values of the other trigonometric functions for angle S, we will use the given information and the definitions of the trigonometric functions.

a) Given cos S = 1/3 and the terminal side of angle S is in Q1.
Since cos S is positive in Q1, we know that S is an acute angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- sin S = √(1 - cos^2 S) = √(1 - (1/3)^2) = √(1 - 1/9) = √(8/9) = √8/3
- tan S = sin S / cos S = (√8/3) / (1/3) = √8
- csc S = 1 / sin S = 1 / (√8/3) = 3/√8 = 3√8/8
- sec S = 1 / cos S = 1 / (1/3) = 3
- cot S = 1 / tan S = 1 / √8 = √8/8

b) Given tan S = -10/7 and the terminal side of S is in Q2.
Since tan S is negative in Q2, we know that S is an obtuse angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- sin S = tan S / √(1 + tan^2 S) = (-10/7) / √(1 + (-10/7)^2) = (-10/7) / √(1 + 100/49) = (-10/7) / √(49/49 + 100/49) = (-10/7) / √(149/49) = (-10/7) / (√149/7) = -10/√149
- cos S = 1 / √(1 + tan^2 S) = 1 / √(1 + (-10/7)^2) = 1 / √(1 + 100/49) = 1 / √(49/49 + 100/49) = 1 / √(149/49) = 1 / (√149/7) = 7/√149
- csc S = 1 / sin S = 1 / (-10/√149) = -√149/10
- sec S = 1 / cos S = 1 / (7/√149) = √149/7
- cot S = 1 / tan S = 1 / (-10/7) = -7/10

c) Given sin S = -5/14 and the terminal side of S is in Q3.
Since sin S is negative in Q3, we know that S is an obtuse angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- cos S = √(1 - sin^2 S) = √(1 - (-5/14)^2) = √(1 - 25/196) = √(196/196 - 25/196) = √(171/196) = √171/14
- tan S = sin S / cos S = (-5/14) / (√171/14) = -5/√171
- csc S = 1 / sin S = 1 / (-5/14) = -14/5
- sec S = 1 / cos S = 1 / (√171/14) = 14/√171
- cot S = 1 / tan S = 1 / (-5/√171) = -√171/5

d) Given sec S = 13/11 and the terminal side of S is in Q4.
Since sec S is positive in Q4, we know that S is an acute angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- cos S = 1 / sec S = 1 / (13/11) = 11/13
- sin S = √(1 - cos^2 S) = √(1 - (11/13)^2) = √(1 - 121/169) = √(169/169 - 121/169) = √(48/169) = √48/13
- tan S = sin S / cos S = (√48/13) / (11/13) = √48/11
- csc S = 1 / sin S = 1 / (√48/13) = 13/√48 = 13√48/48 = 13√3/4
- cot S = 1 / tan S = 1 / (√48/11) = 11/√48 = 11√48/48 = 11√3/4

In summary, for each given case:
a) sin S = √8/3, tan S = √8, csc S = 3√8/8, sec S = 3, cot S = √8/8
b) sin S = -10/√149, cos S = 7/√149, csc S = -√149/10, sec S = √149/7, cot S = -7/10
c) cos S = √171/14, tan S = -5/√171, csc S = -14/5, sec S = 14/√171, cot S = -√171/5
d) sin S = √48/13, cos S = 11/13, tan S = √48/11, csc S = 13√3/4, sec S = 13/11, cot S = 11√3/4