Use identities to find the value of each expression

Question

1. Find Cot theta and Tan theta if (Sin theta=1/3) and (Cot theta>0)
2. Find Csc theta and Sec theta if (Cot theta=-5/2) and (Sin theta>0)

Answer 1

1. When both sin(θ) and cos(θ) are positive, it means θ is in Q1:

Using the Pythagorean Theorem to solve for "x":

x² + 1² = 3²

x² = 8

x = 2√2

cot(θ) = adj/opp = 2√2/1 = 2√2

tan(θ) = opp/adj = 1/(2√2) = √2/4

2. When cot(θ) is negative and sin(θ) is positive, then θ is in Q2:

Using the Pythagorean Theorem to solve for "h":

(-5)² + 2² = h²

h = √29

csc(θ) = hyp/opp = √29/2

sec(θ) = hyp/adj = -√29/5

Answer 2

sinT = 1/3 = opposite side over hypotenuse = .333....

T = arcsin.333....= about 19.47, 180-19.47, or either +360n where n= any integer

cos19.47 = about .943

tan19.47 = about .354

cos(180-19.47)= cos160.53 = -.943

tan160.53 = -.354

cosT= +/-.943

tanT =+/-.354

but there's another restriction cotT>0 which means T is in quadrant 1 or 3

with 0<T<90 or 180<T<270 degrees

so only the positive cosT and positive tanT apply with positive solutions, .943 ad .354

another approach:

sinT=1/3 opposite side=1, hypotenuse =3, adjacent side = sqr(9-1) = sqr8 =2sqr2

cosT = adjacent/hypotenuse = 2sqr2/3= about 2(1.414)/3 = 2.828/3 = .943

tanT =opposite/adjacent , opposite = 1, adjacent = sqr(3^2-1^2) = sqr8

= 1/sqr8 = sqr8/8 = .354

cotT =-5/2, sinT>0 puts T in the 2nd quadrant with 90<T<270

cotT =1/tanT

tanT=1/cotT = 1/(-5/2) =-2/5 =-.4

cscT = hypotenuse/opposite hypotenuse = sqr((-5)^2 +2^2) = sqr29, opposite = 2

= 1/sinT = 1/(sqr(29)/2= 1/2sqr29 = sqr29/2 = about 2.7