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what does 6 mean?

also do you mean tan^2(u) and sin^2(u) because I see tan(2u) and sin(2u)

Also are you asking us to refer to section 5.1 of your book?

I don't see how that is possible.

idk yes my book 6 the number of the problem im on and idk what she means
'

well you are right cos^2+ sin^2 =1 right !
BUT you need to prove( 1+tan^2 ) (1-sin^2i) equal to 1

sohow do i do that

other than that I can't refer to any thing in your book because I don't think I have your book

but i can help you prove the identity above

lol i just need help with it i seriouly dont know what else to do

try to see if you can follow what I said above

divide both sides by cos^2(x) of the identity I spoke about just now
and tell me what you have

i have sin= 1 right idk im horrible at math

Did you try to divide both sides of cos^2(x)+sin^2(x)=1
by cos^2(x)?

yes

so do you know what cos^2(x)/cos^2(x)=?
or what sin^2(x)/cos^2(x)=?
or what 1/cos^2(x)=?

how did you gwt all that

Well I asked you to divide both sides of cos^2(x)+sin^2(x)=1
by cos^2(x)

\[\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{cos^2(x)}\]

ohhh i did it way wrong

\[\frac{\cos^2(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}\]

cos^2(x)/cos^2(x)=1
sin^2(x)/cos^2(x)=tan^2(x)
1/cos^2(x)=sec^2(x)`

though you could just leave it as 1+tan^2(x)=1/cos^2(x) if you want

now plug in both of those results we got from the Pythagorean identity