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.

What's the trouble? Just draw the triangles, label the sides, and read off the trig function values.

If that's too hard, review the function definitions again, post your answers, and then we can see what help you need.

To find the values of the other trigonometric functions (sine, tangent, cosecant, secant, and cotangent) for angle S, we can use the given information about one trigonometric function and the quadrant in which the terminal side of angle S lies. Here's how we can find the values of the other functions for each scenario:

a) Given: cos S = 1/3 and terminal side in Q1

To find the other trigonometric functions, we can use the following relationships for angle S in Q1:
sin S = sqrt(1 - cos^2 S)
tan S = sin S / cos S
csc S = 1 / sin S
sec S = 1 / cos S
cot S = 1 / tan S

Plugging in the value cos S = 1/3 in the above equations, we get:
sin S = sqrt(1 - (1/3)^2) = sqrt(8/9) = 2sqrt(2)/3
tan S = (2sqrt(2)/3) / (1/3) = 2sqrt(2)
csc S = 1 / (2sqrt(2)/3) = 3sqrt(2)/2
sec S = 1 / (1/3) = 3
cot S = 1 / (2sqrt(2)) = sqrt(2)/4

Therefore, the values of the other trigonometric functions for angle S in this scenario are:
sin S = 2sqrt(2)/3
tan S = 2sqrt(2)
csc S = 3sqrt(2)/2
sec S = 3
cot S = sqrt(2)/4

b) Given: tan S = -10/7 and terminal side in Q2

In Q2, tangent is negative, so we have:
sin S = -sqrt(1 / (1 + tan^2 S))
cos S = 1 / sqrt(1 + tan^2 S)

Plugging in the value tan S = -10/7 in the above equations, we get:
sin S = -sqrt(1 / (1 + (-10/7)^2)) = -sqrt(147) / 17
cos S = 1 / sqrt(1 + (-10/7)^2) = 3 / 17

To find the other trigonometric functions, we can use the reciprocals and ratios derived from the sin S and cos S values:
csc S = 1 / sin S
sec S = 1 / cos S
cot S = 1 / tan S

Plugging in the sin S and cos S values, we get:
csc S = 1 / (-sqrt(147) / 17) = -17 / sqrt(147)
sec S = 1 / (3 / 17) = 17 / 3
cot S = 1 / (-10/7) = -7/10

Therefore, the values of the other trigonometric functions for angle S in this scenario are:
sin S = -sqrt(147) / 17
cos S = 3 / 17
csc S = -17 / sqrt(147)
sec S = 17 / 3
cot S = -7/10

c) Given: sin S = -5/14 and terminal side in Q3

In Q3, sine is negative, so we have:
cos S = -sqrt(1 / (1 + sin^2 S))
tan S = sin S / cos S

Plugging in the value sin S = -5/14 in the above equations, we get:
cos S = -sqrt(1 / (1 + (-5/14)^2)) = -3sqrt(119) / 14
tan S = (-5/14) / (-3sqrt(119)/14) = 5 / (3sqrt(119))

To find the other trigonometric functions, we can use the reciprocals and ratios derived from the sin S and cos S values:
csc S = 1 / sin S
sec S = 1 / cos S
cot S = 1 / tan S

Plugging in the sin S and cos S values, we get:
csc S = 1 / (-5/14) = -14/5
sec S = 1 / (-3sqrt(119) / 14) = -14 / (3sqrt(119))
cot S = 1 / (5 / (3sqrt(119))) = 3sqrt(119) / 5

Therefore, the values of the other trigonometric functions for angle S in this scenario are:
sin S = -5/14
cos S = -3sqrt(119) / 14
csc S = -14/5
sec S = -14 / (3sqrt(119))
cot S = 3sqrt(119) / 5

d) Given: sec S = 13/11 and terminal side in Q4

In Q4, secant is positive, so we have:
cos S = 1 / sec S
sin S = sqrt(1 - cos^2 S)

Plugging in the value sec S = 13/11 in the above equations, we get:
cos S = 1 / (13/11) = 11/13
sin S = sqrt(1 - (11/13)^2) = 2sqrt(6)/13

To find the other trigonometric functions, we can use the reciprocals and ratios derived from the sin S and cos S values:
tan S = sin S / cos S
csc S = 1 / sin S
cot S = 1 / tan S

Plugging in the sin S and cos S values, we get:
tan S = (2sqrt(6)/13) / (11/13) = 2sqrt(6)/11
csc S = 1 / (2sqrt(6)/13) = 13 / (2sqrt(6))
cot S = 1 / (2sqrt(6)/11) = 11 / (2sqrt(6))

Therefore, the values of the other trigonometric functions for angle S in this scenario are:
sin S = 2sqrt(6)/13
cos S = 11/13
tan S = 2sqrt(6)/11
csc S = 13 / (2sqrt(6))
sec S = 13/11
cot S = 11 / (2sqrt(6))

I hope this explanation helps! Let me know if you have any further questions.