How do you find the five remaining trigonometric function satisfying sectheta=6/5, tantheta<0?

2 Answers
Gerardina C.
Jun 20, 2017

costheta=5/6; sintheta=-sqrt11/6; tantheta=-sqrt11/5; cottheta=-5/sqrt11; csctheta=-6/sqrt11

Explanation:

Since sectheta=1/costheta=6/5, you get

costheta=1/sectheta=5/6

Since

sintheta=+-sqrt(1-cos^2theta),

tantheta<0,

costheta>0,

tantheta=sintheta/costheta,

then

sintheta=-sqrt(1-cos^2theta)=-sqrt(1-25/36)=-sqrt11/6

and

tantheta=(-sqrt11/cancel6)/(5/cancel6)=-sqrt11/5

cottheta=1/tantheta=-5/sqrt11

csctheta=1/sintheta=-6/sqrt11

EZ as pi
Jun 20, 2017

sin theta" "cos theta" "tan theta" "cot theta" "sec theta" cosec "theta"

(-sqrt11)/6" "5/6" "(-sqrt11)/5" "5/(-sqrt11)" "6/5" "6/(-sqrt11)

Explanation:

sec theta = 6/5 =r/x

Therefore we can find y using Pythagoras' Theorem.

y = sqrt(6^2-5^2) = sqrt11

We need to decide which quadrant we are in.

sec theta is given as positive rarr 1st or 4th
tan theta is given as negative rarr 2nd or 4th

So the only quadrant where both condition hold is the 4th

In the 4th quadrant, only cos theta and sec theta are positive.

We can now write all 6 ratios:

x=5, " "y =-sqrt11" " and " "r =6

sin theta" "cos theta" "tan theta" "cot theta" "sec theta" cosec "theta"

-y/r" "x/r" "-y/x" "-x/y" "r/x" "-r/y"

(-sqrt11)/6" "5/6" "(-sqrt11)/5" "5/(-sqrt11)" "6/5" "6/(-sqrt11)

Related questions
See all questions in Applying Trig Functions to Angles of Rotation
Impact of this question
36289 views around the world
You can reuse this answer
Creative Commons License