Verify the identity. Show your work. 1 + sec2xsin2x = sec2x the answer is 2

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Verify the identity. Show your work. 1 + sec2xsin2x = sec2x the answer is 2

Mathematics
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Teacher:Substitute for the sec^2x on the RHS only. This is a reciprocal function. You should be able to take this from here. Me: wtf?

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I think somebody already answered this oooover here http://openstudy.com/study#/updates/509f00d7e4b013fc35a19250
i used that one but my teacher said that up there so im confuzled
@Vocaloid can you give me some understanding of my teachers words plz
sec^2(x) = 1/(cos^2(x))
that's basically what she's saying, substitute 1/(cos^2(x)) for sec^2(x) on the right side
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this is to confusing this is why i hate math lol
That's why I'm helping you. :D
Do you know how to multiply this?
true that thnx alot lol
i really dont
For instance, what is \(4\times \frac{ 1 }{ 3}=?\) You only need to put 4 in the numerator. So you'll get \(\frac{ 4 }{ 3 }\).
Now, what is \( 1 + (\frac{ 1 }{ \cos^2x }\times \sin^2x)\)?
hold on ups is at my gate
imma go see what it is
imm home alone so i have to get it smh lazy people deez days i leave my gate open for a reason haha
@mathway im back now
then answer my question now
frac(cos^2(x))
is that right @mathway
??
HI!!
it is clear that \[\frac{\sin(x)}{\cos(x)}=\tan(x)\]?
hiii
yes
and also it is clear that \[\sec(x)=\frac{1}{\cos(x)}\]right?
yess
that means \[\sec^2(x)\sin^2(x)=\frac{\sin^2(x)}{\cos^2(x)}=\tan^2(x)\]
ok
so you are looking at \[1+\tan^2(x)=\sec^2(x)\] which is definitely true, we can get it in one step
start with the mother of all trig identities \[\cos^2(x)+\sin^2(x)=1\] divide both sides by \(\cos^2(x)\) and you get \[\frac{\cos^2(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}\] or \[1+\tan^2(x)=\sec^2(x)\] as needed
woow thats alot to take in lol but it makes since
\[\color\magenta\heartsuit\]

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