## kengeta Group Title how do i prove : tan^2 θ cos^2 θ + cos^2 θ = 1 one month ago one month ago

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1. myininaya Group Title

well my first attempt to prove this is true is to take the left hand side and try to show the right hand side the left hand side has both terms with the factor cos^2(theta) start by factoring cos^2(theta) out from both terms this will get us closer to writting as 1 term just as the right hand side is (one term)

2. myininaya Group Title

let me know if you still don't know where to go after that

3. kengeta Group Title

I'm pretty confused

4. myininaya Group Title

on how to factor?

5. myininaya Group Title

do you know hot to factor the expression ax+x?

6. myininaya Group Title

how* (not hot)

7. kengeta Group Title

no

8. myininaya Group Title

I will show you how to factor ax+x then you should be able to factor tan^2(theta)*cos^2(theta)+cos^2(theta) so ax+x I see there is a x in both terms |dw:1407522006591:dw| so i will factor that x out like so $ax+1x=x(a+1)$

9. myininaya Group Title

you have $\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1$ $\text{ do you see that there is a } cos^2(\theta) \text{ in both terms ?}$

10. kengeta Group Title

yes

11. myininaya Group Title

can you factor that out using what you know about multiplication and division like i did above?

12. myininaya Group Title

also this way i'm asking you to approach it is not the only approach to take

13. myininaya Group Title

ok if you are having trouble factoring the cos^2(theta) out how about we try this another way do you recall tan is sin/cos?

14. kengeta Group Title

yes

15. myininaya Group Title

$\frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta)$ in the first term there, do you see anything that cancels?

16. kengeta Group Title

cos^2?

17. myininaya Group Title

right so what does that leave us with?

18. myininaya Group Title

$\frac{\sin^2(\theta)}{\cancel{\cos^2(\theta)}}\cancel{\cos^2(\theta)}+\cos^2(\theta)$ what does this give us?

19. kengeta Group Title

sin^2(theta)+cos^2(theta)

20. myininaya Group Title

And that equals?

21. myininaya Group Title

you need to recall some trig identities for these trigonometric proofs

22. myininaya Group Title

This one of the most basic ones It actually has a name

23. kengeta Group Title

okay si... 1?

24. myininaya Group Title

I will give you a hint: Pythagorean Identity

25. myininaya Group Title

yep

26. kengeta Group Title

*so

27. kengeta Group Title

okay i get it now thanks soooo much

28. myininaya Group Title

Way 1: $\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1= \\ \cos^2(\theta)(\tan^2(\theta)+1)= \\ \cos^2(\theta) \cdot \sec^2(\theta)= \\ 1$ Way 2: $\tan^2(\theta) \cos^2(\theta)+\cos^2(\theta)= \\ \frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta) = \\ \sin^2(\theta) +\cos^2(\theta)= \\ 1$ These are the two ways that I can think of... This does not mean they are the only ways.

29. myininaya Group Title

You don't have to do both ways. Just one way.

30. myininaya Group Title

I used a Pythagorean identity in both ways.

31. myininaya Group Title

way 1 was the way I was trying to get you to go about it the first time around

32. myininaya Group Title

I don't think one way is more harder than the other But you will have to review some algebra especially if you don't remember how to factor. Factoring will come up again.