Is sec Θ ° > tan Θ ° for every angle?

Question

Please explain why or why not?

Answer 1

Find a positive number x for which cos (ln x) = 0.

I really don't know where to start, please help me.

Answer 2

let's look at their graphs

http://www.wolframalpha.com/input/?i=y+%3D+1%2Fcosx+%2C+y+%3D+tanx

for some values, the secant curve is above the tangent curve, for others the opposite

easiest way is to look at the CAST rule
in quad IV , the secant is positive, but the tangent is negative, so secØ > tanØ
in quad III, the secant is negative, but the tangent is positive, so secØ < tanØ

e.g. Ø = 315°
sec 315° = √2
tan45 = -1
sec45 > tan45 in IV

Ø = 225°
sec225 = -√2 = -1.414..
tan225 = 1
thus tanØ > secØ in III

you second question....

If cos(lnx) = 0
then lnx could be π/2 , since we know cos(π/2) = 0

so if ln(x) = π/2
then x = appr 4.8105

check:
set your calculator to radians
take ln(4.8105)
now press 'cos' to get close to zero

Answer 3

may be easier to recognize that

sec = 1/cos
tan = sin/cos

sec > tan implies

1/cos > sin/cos

which is true iff

cos > 0

that's in quadrants 1 and 4

Answer 4

Thank you both, reiny and jolly rancher.

Thanks for the graph, and jolly rancher, that totally made it a lot easier for me to understand. :)