What is the exact value of the other primary trigonometric ratios of #tanTheta=(-5)/12# in quadrant II?

1 Answer
Apr 7, 2018

#cot(theta) = -12/5#

#sec(theta) = -13/12#

#cos(theta) = -12/13#

#sin(theta) = 5/13#

#csc(theta) =13/5#

Explanation:

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

#cot(theta) = -12/5#

Use the identity:

#1+tan^2(theta) = sec^2(theta)#

Substitute #tan^2(theta) = (-5/12)^2 #:

#1+(-5/12)^2 = sec^2(theta)#

#144/144+25/144#

#169/144 = sec^2(theta)#

#sec(theta) = +-13/12#

We know that the secant is negative in the second quadrant, therefore, we choose the negative value:

#sec(theta) = -13/12#

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

#cos(theta) = -12/13#

Use the identity:

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

Multiply both sides by #cos(theta)#:

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

Substitute #tan(theta) = -5/12 and cos(theta) = -12/13#:

#sin(theta) = (-5/12)(-12/13)#

#sin(theta) = 5/13#

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

#csc(theta) = 13/5#