Question

Verify the identity. Show your work.
(1 + tan2u)(1 - sin2u) = 1
my answer > cos^2x+sin^2x=1

Answer 1

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.

Answer 2

@phi @Vocaloid

Answer 3

what does 6 mean?

Answer 4

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

Answer 5

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

Answer 6

I don't see how that is possible.

Answer 7

I am with @freckles It must be \((1+tan^2u)(1-sin^2u) =1\)

Answer 8

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

Answer 9

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

Answer 10

sohow do i do that

Answer 11

\[\huge\rm (1+\tan^2u)(1-\sin^2u)=1\]
show that R.H.S=L.H.S
are u familiar with the trig identities you have to apply that :=)

Answer 12

\[\cos^2(x)+\sin^2(x)=1 \\ \text{ subtract } \sin^2(x) \text{ on both sides } \\ \cos^2(x)=1-\sin^2(x) \\ \text{ hint: now divide both sides of } \cos^2(x)+\sin^2(x)=1 \\ \text{ by } \cos^2(x)\]

Answer 13

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

Answer 14

but i can help you prove the identity above

Answer 15

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

Answer 16

try to see if you can follow what I said above

Answer 17

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

Answer 18

i have sin= 1 right idk im horrible at math

Answer 19

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

Answer 20

yes

Answer 21

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)=?

Answer 22

how did you gwt all that

Answer 23

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

Answer 24

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

Answer 25

ohhh i did it way wrong

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

\(1+tan^2u = sec^2 u =\dfrac{1}{cos^2u}\\1-sin^2 u = cos^2u\\hence,(1+tan^2u)(1-sin^2u)= \dfrac{1}{\cancel{cos^2u}}*\cancel{cos^2u}=1\)