Question #93462

Question

2054 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-30T22:26:31

#tan(theta) = 5/3#
#sec(theta) = -sqrt(34)/3#
#cos(theta) = -(3sqrt(34))/34#
#sin(theta) = -(5sqrt(34))/34#
#csc(theta) = -sqrt(34)/5#

Given: #cot(theta) = 3/5 and sec(theta) < 0#

Use the identity #tan (theta) = 1/cot(theta)#:

#tan(theta) = 1/(3/5)#

#tan(theta) = 5/3#

Use the identity #1 + tan^2(theta) = sec^2(theta)#

#1 + (5/3)^2 = sec^2(theta)#

#sec^2(theta) = 34/9#

#sec(theta) = -sqrt(34)/3#

Use the identity #cos(theta) = 1/sec(theta)#

#cos(theta) = -3/sqrt(34)#

#cos(theta) = -(3sqrt(34))/34#

Use the identity #tan(theta) = sin(theta)/cos(theta)#

#tan(theta)cos(theta) = sin(theta)#

#sin(theta) = 5/3(-(3sqrt(34))/34)#

#sin(theta) = (-5sqrt(34))/34#

Use the identity #csc(theta) = 1/sin(theta)#

#csc(theta) = 1/(-5/sqrt(34)#

#csc(theta) = -sqrt(34)/5#