Question

please help question attached

Answer 1

@whpalmer4 or @Mertsj

Answer 2

write in terms of sin theta and cos theta \[\sec \theta -\tan \theta \sin\]

Answer 3

Okay, do you know the definitions of \(\sec \theta\) and \(\tan \theta\) in terms of the elementary trig functions \(\sin\theta\) and \(\cos\theta\)?

Answer 4

yes well sec is the reciprocal of cos is that what u mean ?

Answer 5

Okay, that's a start. What about \(\tan\theta\)?

Answer 6

do i set it up like this
\[\frac{ 1 }{ \cos \theta }-\frac{ \sin \theta }{ \cos \theta }\times \frac{ \sin \theta }{ 1 }\]
@whpalmer4

Answer 7

And so you have \[\frac{ 1-\sin^2\theta }{ \cos \theta }\]

Does the numerator look familiar?

Answer 8

yes - but my answer in the book says the answer is just \[\cos \theta \]

Answer 9

I'll give you a hint.
sin^2x + cos^2x = 1

Answer 10

well is that suppose to be \[\sin^2\theta + \cos^2\theta = 1 \]
that is a pythagorean identity but how does the answer just say cos theta?

Answer 11

same exact thing. Anyhow, manipulate that trigonometric identity into the form 1-sin^2theta

Answer 12

well if i multiply by cos theta i get that right

Answer 13

from what i have above as
\[\frac{ 1-\sin \theta }{ \cos \theta }\]

Answer 14

You don't have to multiply. You have: \[\huge \frac{ 1-\sin^2\theta }{ \cos \theta }\]
and \[\huge \sin^2\theta + \cos^2\theta = 1\]
If we subtract sin^2 from both sides we get:

\[\huge \cos^2\theta = 1 -\sin^2\theta\]

Answer 15

so how is that just cos theta?

Answer 16

Because \[\huge \frac{ \cos^2\theta }{ \cos\theta }=\frac{ \cos\theta*\cos\theta }{ \cos\theta }= \cos\theta\]

Answer 17

or you can just use the fact that \[\large \frac{ a^m }{ a^n }= a^{m-n}\]

Answer 18

says in my book just simplify if possible from what i have not sure how i go to from the 1-sin theta/cos theta ?

Answer 19

\[\large \frac{ 1 }{ \cos \theta }-\frac{ \sin \theta }{ \cos \theta }\times \frac{ \sin \theta }{ 1 } = \frac{ 1 }{ \cos \theta }-\frac{ \sin^2 \theta }{ \cos \theta } = \frac{ 1-\sin^2\theta }{\cos\theta }\]

Answer 20

right got that - so now how does it get just cos theta as the answer from that point

Answer 21

I explained earlier. using sin^2x + cos^2x=1

We have cos^2x = 1-sin^2x which is the numerator on the last fraction

btw give palmer that medal as I'm answer sniping

Answer 22

ok well in the botton of denomitator we have cos theta not cos^2 theta

Answer 23

so you divide cos^2 by cos and you get cos theta

Answer 24

As @bibby said:
\[\large \frac{ 1 }{ \cos \theta }-\frac{ \sin \theta }{ \cos \theta }\times \frac{ \sin \theta }{ 1 } = \frac{ 1 }{ \cos \theta }-\frac{ \sin^2 \theta }{ \cos \theta } = \frac{ 1-\sin^2\theta }{\cos\theta }\]We know that \[\sin^2\theta + \cos^2\theta = 1\]If we subtract \(\sin^2\theta\) from both sides of that, we get\[\sin^2\theta -\sin^2\theta + \cos^2\theta = 1-\sin^2\theta\]\[cos^2\theta = 1-\sin^2\theta\]Substitute that into our previous expression:
\[\frac{1-\sin^2\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \frac{\cos\theta\cancel{*\cos\theta}}{\cancel{\cos\theta}}=\cos\theta\]

Answer 25

oh ok guess just didnt know could do that lol

Answer 26

As I recommended in another question a few moments ago, make yourself a set of trig identity flashcards and drill with them for a few minutes every day. Preferably, make them in various arrangements, too: don't just have one that says \[\sin^2\theta + \cos^2\theta = 1\]but also have one that has \[\sin^2\theta = 1-\cos^2\theta\]and one that says \[\cos^2\theta = 1-\sin^2\theta\]

Answer 27

thanks both for the help guess didnt see it and didnt know u could do that haha thanks @whpalmer4 and @bibby

Answer 28

np np

Answer 29

You're welcome! If you need a nice list of trig identities, I like this one:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html