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

1. 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.

2. anonymous

@phi @Vocaloid

3. freckles

what does 6 mean?

4. freckles

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

5. freckles

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

6. freckles

I don't see how that is possible.

7. Loser66

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

8. anonymous

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

9. 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

10. anonymous

sohow do i do that

11. 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 :=)

12. freckles

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

13. freckles

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

14. freckles

15. anonymous

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

16. freckles

try to see if you can follow what I said above

17. freckles

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

18. anonymous

i have sin= 1 right idk im horrible at math

19. freckles

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

20. anonymous

yes

21. 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)=?

22. anonymous

how did you gwt all that

23. freckles

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

24. freckles

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

25. anonymous

ohhh i did it way wrong

26. freckles

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

27. freckles

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

28. freckles

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

29. freckles

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

30. 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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