Question

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

Answer 1

`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`

Answer 2

`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)`