Question

Given `sec x = 6` for `x` in Q1, what are the values of all the six trigonometric functions?

Answer 1

`{ (sin x = sqrt(35)/6), (cos x = 1/6), (tan x = sqrt(35)), (csc x = (6sqrt(35))/35), (sec x = 6), (cot x = sqrt(35)/35) :}`

Explanation:

Note that:

`sec x = 1/cos x`

So:

`cos x = 1/sec x = 1/6`

By Pythagoras, we know that:

`cos^2 x + sin^2 x = 1`

Hence:

`sin x = +-sqrt(1-cos^2 x) = +-sqrt(1-(1/6)^2) = +-sqrt(35/36) = +-sqrt(35)/6`

We are told that `sin x > 0`, so `sin x = sqrt(35)/6`

Then

`csc x = 1/sin x = 6/sqrt(35) = (6sqrt(35))/35`

Then:

`tan x = sin x / cos x = (sqrt(35)/6) / (1/6) = sqrt(35)`

`cot x = 1/tan x= 1/sqrt(35) = sqrt(35)/35`

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Alternatively, just think of a right angled triangle with sides:

`"adjacent" = 1`

`"opposite" = sqrt(35)`

`"hypotenuse" = 6`

since `1^2+sqrt(35)^2 = 1+35 = 36 = 6^2`

Then:

`sin x = "opposite"/"hypotenuse" = sqrt(35)/6`

`cos x = "adjacent"/"hypotenuse" = 1/6`

`tan x = "opposite"/"adjacent" = sqrt(35)/1 = sqrt(35)`

`csc x = "hypotenuse"/"opposite" = 6/sqrt(35) = (6sqrt(35))/35`

`sec x = "hypotenuse"/"adjacent" = 6/1 = 6`

`cot x = "adjacent"/"opposite" = 1/sqrt(35) = sqrt(35)/35`