## UnbelievableDreams one year ago Confusing math question

1. anonymous

2. UnbelievableDreams

Choose the correct expression that completes the identity. sec^2 (x) - tan^2(x)

3. anonymous

im sorry this is above my math intelligence

4. Nnesha

sec^2 = what identity ?

5. Nnesha

do you have sin ,cos identities notes ?

6. dan815

what are the choices

7. dan815

|dw:1438637947361:dw|

8. dan815

|dw:1438637971118:dw|

9. UnbelievableDreams

No, I don't have notes. I am so confused.

10. Nnesha

now u got the answer if you apply identities that would be easy $\huge\rm tan^2 +1=\sec^2$

11. Nnesha

replace sec^2 by tan^2 +1 $\huge\rm \color{reD}{sec^2(x)}-\tan^2(x)$ $\huge\rm \color{reD}{tan^2+1 }-\tan^2$

12. dan815

u need to know only the basic ones sin^2(x)+cos^2(x)=1 and sec(x)=1/cos(x) csc(x)=1/sin(x) cot(x)=1/tan(x) there are the basic ones to know

13. UnbelievableDreams

I am still confused. :(

14. freckles

do you know the Pythagorean identity(ies)?

15. freckles

$\sin^2(x)+\cos^2(x)=1 \text{ do you know the one mentioned by dan ?}$

16. UnbelievableDreams

No.

17. freckles

Do you know how to graph the equation x^2+y^2=1? This is the circle with center (0,0) and radius 1. Looks like this: |dw:1438638836144:dw| say we pick a random point (a,b) on the circle: |dw:1438638871637:dw| now we draw a right triangle |dw:1438638917068:dw| we know the relationship between the side of those right triangle to be $a^2+b^2=1^2 \\ \text{ or } \\a^2+b^2=1 \\ \text{ the 1 comes from the radius of the circle being 1 }$ now we could write this same equation in terms of theta by using the drawing

18. freckles

$\cos(\theta)=\frac{a}{1} \text{ or } \cos(\theta)=a \\ \sin(\theta)=\frac{b}{1} \text{ or } \sin(\theta)=b \\ \text{ squaring both sides gives } \\ (\cos(\theta))^2=a^2 \text{ also written as } \cos^2(\theta)=a^2 \\ (\sin(\theta))^2=b^2 \text{ also written as } \sin^2(\theta)=b^2 \\ \text{ so replacing the items \in } a^2+b^2=1 \\ \text{ we have } \cos^2(\theta)+\sin^2(\theta)=1$

19. freckles

you can get the identity that nnesha mentioned from this if you divide both sides by cos^2(theta)

20. freckles

$(\cos(\theta))^2+(\sin(\theta))^2=1 \\ \text{ dividing both sides by } (\cos(\theta))^2 \\ \frac{(\cos(\theta))^2+(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1}{(\cos(\theta))^2} \\ \text{ separating fraction on \left hand side } \\ \frac{(\cos(\theta))^2}{(\cos(\theta))^2}+\frac{(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1^2}{(\cos(\theta))^2} \text{ note: notice I replace } 1 \text{ with } 1^2 \\ \text{ now using law of exponents } \\ (\frac{\cos(\theta)}{\cos(\theta)})^2+(\frac{\sin(\theta)}{\cos(\theta)})^2=(\frac{1}{\cos(\theta)})^2 \\ \text{ now there were ratios mentioned by dan that were equivalent \to some you see here }$ for example you know $\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta) \\ \text{ and } \frac{1}{\cos(\theta)}=\sec(\theta)$

21. freckles

$\text{ also you know } \frac{\cos(\theta)}{\cos(\theta)}=1 \text{ on it's domain } \\ \text{ so we have } 1^2+(\tan(\theta))^2=(\sec(\theta))^2 \\\ \text{ which can be written as } \\ 1+\tan^2(\theta)=\sec^2(\theta)$

22. freckles

this is the Pythagorean identity you want to apply

23. UnbelievableDreams

Ah, I see.

24. UnbelievableDreams

Can you help me another one?

25. freckles

I can try

26. UnbelievableDreams

It is similar to this question. But I like to see step by step. 1-tan^2x= ___________

27. freckles

do you want to write it in terms of sec or use difference of squares

28. freckles

$a^2-b^2=(a-b)(a+b) \text{ is difference of squares in factored form }$

29. freckles

if you want to write it in terms of sec you can use the identity we just mentioned $1+\tan^2(\theta)=\sec^2(\theta) \\ \text{ subtract } 1 \text{ on both sides } \\ \tan^2(\theta)=\sec^2(\theta)-1 \\ \text {replace } \tan^2(\theta) \text{ with } \sec^2(\theta)-1$

30. UnbelievableDreams

31. UnbelievableDreams

How do you get sec?

32. freckles

did you replace tan^2(theta) with sec^2(theta)-1 ?

33. freckles

$1-\tan^2(\theta) \\ \text{ replace } \tan^2(\theta) \text{ with } \sec^2(\theta)-1 \\ 1-(\sec^2(\theta)-1))$

34. freckles

distribute and combine like terms but this shouldn't give you 1/sec^2(x)

35. UnbelievableDreams

Sec^2x - 1 ?

36. freckles

nope you distribute $1-\tan^2(x) \\ \text{ replace } \tan^2(x) \text{ with } \sec^2(x)-1 \\ \text{ which gives } \\ 1-(\sec^2(x)-1) \\ \text{ now distribute } \\1-\sec^2(x)+1 \\ \text{ combine like terms } (1+1)-\sec^2(x)$

37. UnbelievableDreams

Ah, I see.