Question

help with secx/sinx-sinx/cosx!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

you wanna simplify it?
note that sec x = 1/cos x
so it becomes \(\frac{1}{sin x cos x} - \frac{sin x}{cos x} \)
if you continue to put both fractions into one , you'll notice something beautiful.

Answer 2

well i got ((1/cosx)/sinx)-sinx/cos x and IDK what what to do after that unless i change cosx to 1/secx then can i cancel out the two secx?

Answer 3

The sec x cannot be "cancelled".
from above, \[\frac{ 1 }{ \sin x \cos x }-\frac{ \sin x }{ \cos x }=\frac{ 1-\sin x(\sin x) }{ \sin x \cos x}\]
and \(cos^2 x+sin^2 x =1 \)
:) I think this will help.

Answer 4

but if you subtract sin from 1 how did you get 1-sinx(sinx)? wouldn't it just be 1-sinx?

Answer 5

hmm...I didn't subtract sin x..
\[\frac{ 1 }{ a }-\frac{ 1 }{ b }=\frac{ b-a }{ ab }\]
I think this should clear things up.

Answer 6

but you have two things in the first one cuz it's1/sinxcosx not just 1/sinx

Answer 7

yup! :) so I multiplied sin x to \( \frac {sin x}{cos x}\) to have the same denominator.

Answer 8

oh okay then you seperated it into 1-sinx(sinx) because when you multiply it you get 1-sin^2x

Answer 9

but now what? cuz i know it has to equal cotx but i don't see how

Answer 10

well, since from \(cos^2x+sin^2x=1\) we know that \(cos^2 x=1-sin^2 x\)
so it simplifies to \(\frac{cos^2}{sin x cos x}\)=cot x

Answer 11

but there's a cosx in the denominator so how does that work if cot x =cosx/sinx?

Answer 12

well, \( \frac{cos^2 x}{ sinxcosx}\)= \(\frac{cos x cosx}{sin x cos x}\) and we can cancel the cos x from the numerator and denominator.

Answer 13

oh okay!!!! weLl thanx

Answer 14

You're welcome :)