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OpenStudy (anonymous):
see secx*cosx=1
OpenStudy (luigi0210):
distribute!
Parth (parthkohli):
\[\sec(x) = \dfrac{1}{\cos(x)}\]So\[\dfrac{1}{\cos(x)}\left(1 + \cos(x)\right) = 1 + \dfrac{1}{\cos(x)}\]Distribute\[\dfrac{1}{\cos(x)} + 1 = 1 + \dfrac{1}{\cos(x)}\]Don't need to go further
Parth (parthkohli):
The left hand side is just the right hand side, right?
OpenStudy (anonymous):
Presumably
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Parth (parthkohli):
So we're done.
Parth (parthkohli):
Could be shorter by memorizing \(\sec(x) \times \cos(x) = 1\), but bleh :p
OpenStudy (anonymous):
wouldn't you prove it like:
secx+secx*cosx
=secx+1/cosx*cosx
=secx+1
Parth (parthkohli):
Yeah, that's the shorter way to prove it.
OpenStudy (anonymous):
:D
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OpenStudy (jhannybean):
can you elaborate @isuckatmath9999 ? :)i mean elaborate your way. haha
OpenStudy (zzr0ck3r):
lol there are about 5000* ways, anyone else got another?
Parth (parthkohli):
@Jhannybean \(\sec(x) \times \cos(x) = 1\)
OpenStudy (anonymous):
I don't neeed another its for an assignment lol. :)
OpenStudy (jhannybean):
yeahhh but how'd he get...sec (x) + 1/cos^2 (x)...
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OpenStudy (jhannybean):
OH YOU FACTORED OUT THE SEC (X)???
OpenStudy (anonymous):
yea lol
Parth (parthkohli):
It's not 1/cos^2(x) lol
It's 1/cos(x) * cos(x)
OpenStudy (jhannybean):
oh ho ho, i gotchu.
OpenStudy (anonymous):
I have more questions to debate upon lol. don't worry :P heheee
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