anonymous
  • anonymous
Verify the identity. Show your work. (1 + tan2u)(1 - sin2u) = 1 my answer > cos^2x+sin^2x=1
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
but my teacher said 6. You can re-write each factor as another trig function squared. 1+tan^2u=(standard identity)? and 1-sin^2u=(standard identity)? Once you do that, the next steps are clear. Please refer to section 5.1 of your text for examples and sample problems.
anonymous
  • anonymous
@phi @Vocaloid
freckles
  • freckles
what does 6 mean?

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freckles
  • freckles
also do you mean tan^2(u) and sin^2(u) because I see tan(2u) and sin(2u)
freckles
  • freckles
Also are you asking us to refer to section 5.1 of your book?
freckles
  • freckles
I don't see how that is possible.
Loser66
  • Loser66
I am with @freckles It must be \((1+tan^2u)(1-sin^2u) =1\)
anonymous
  • anonymous
idk yes my book 6 the number of the problem im on and idk what she means '
Nnesha
  • Nnesha
well you are right cos^2+ sin^2 =1 right ! BUT you need to prove( 1+tan^2 ) (1-sin^2i) equal to 1
anonymous
  • anonymous
sohow do i do that
Nnesha
  • Nnesha
\[\huge\rm (1+\tan^2u)(1-\sin^2u)=1\] show that R.H.S=L.H.S are u familiar with the trig identities you have to apply that :=)
freckles
  • freckles
\[\cos^2(x)+\sin^2(x)=1 \\ \text{ subtract } \sin^2(x) \text{ on both sides } \\ \cos^2(x)=1-\sin^2(x) \\ \text{ hint: now divide both sides of } \cos^2(x)+\sin^2(x)=1 \\ \text{ by } \cos^2(x)\]
freckles
  • freckles
other than that I can't refer to any thing in your book because I don't think I have your book
freckles
  • freckles
but i can help you prove the identity above
anonymous
  • anonymous
lol i just need help with it i seriouly dont know what else to do
freckles
  • freckles
try to see if you can follow what I said above
freckles
  • freckles
divide both sides by cos^2(x) of the identity I spoke about just now and tell me what you have
anonymous
  • anonymous
i have sin= 1 right idk im horrible at math
freckles
  • freckles
Did you try to divide both sides of cos^2(x)+sin^2(x)=1 by cos^2(x)?
anonymous
  • anonymous
yes
freckles
  • freckles
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)=?
anonymous
  • anonymous
how did you gwt all that
freckles
  • freckles
Well I asked you to divide both sides of cos^2(x)+sin^2(x)=1 by cos^2(x)
freckles
  • freckles
\[\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{cos^2(x)}\]
anonymous
  • anonymous
ohhh i did it way wrong
freckles
  • freckles
\[\frac{\cos^2(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}\]
freckles
  • freckles
cos^2(x)/cos^2(x)=1 sin^2(x)/cos^2(x)=tan^2(x) 1/cos^2(x)=sec^2(x)`
freckles
  • freckles
though you could just leave it as 1+tan^2(x)=1/cos^2(x) if you want
freckles
  • freckles
now plug in both of those results we got from the Pythagorean identity
Loser66
  • Loser66
\(1+tan^2u = sec^2 u =\dfrac{1}{cos^2u}\\1-sin^2 u = cos^2u\\hence,(1+tan^2u)(1-sin^2u)= \dfrac{1}{\cancel{cos^2u}}*\cancel{cos^2u}=1\)

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