anonymous
  • anonymous
how do i prove : tan^2 θ cos^2 θ + cos^2 θ = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
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)
myininaya
  • myininaya
let me know if you still don't know where to go after that
anonymous
  • anonymous
I'm pretty confused

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myininaya
  • myininaya
on how to factor?
myininaya
  • myininaya
do you know hot to factor the expression ax+x?
myininaya
  • myininaya
how* (not hot)
anonymous
  • anonymous
no
myininaya
  • myininaya
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 \[ax+1x=x(a+1)\]
myininaya
  • myininaya
you have \[\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1 \] \[\text{ do you see that there is a } cos^2(\theta) \text{ in both terms ?}\]
anonymous
  • anonymous
yes
myininaya
  • myininaya
can you factor that out using what you know about multiplication and division like i did above?
myininaya
  • myininaya
also this way i'm asking you to approach it is not the only approach to take
myininaya
  • myininaya
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?
anonymous
  • anonymous
yes
myininaya
  • myininaya
\[\frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta)\] in the first term there, do you see anything that cancels?
anonymous
  • anonymous
cos^2?
myininaya
  • myininaya
right so what does that leave us with?
myininaya
  • myininaya
\[\frac{\sin^2(\theta)}{\cancel{\cos^2(\theta)}}\cancel{\cos^2(\theta)}+\cos^2(\theta)\] what does this give us?
anonymous
  • anonymous
sin^2(theta)+cos^2(theta)
myininaya
  • myininaya
And that equals?
myininaya
  • myininaya
you need to recall some trig identities for these trigonometric proofs
myininaya
  • myininaya
This one of the most basic ones It actually has a name
anonymous
  • anonymous
okay si... 1?
myininaya
  • myininaya
I will give you a hint: Pythagorean Identity
myininaya
  • myininaya
yep
anonymous
  • anonymous
*so
anonymous
  • anonymous
okay i get it now thanks soooo much
myininaya
  • myininaya
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.
myininaya
  • myininaya
You don't have to do both ways. Just one way.
myininaya
  • myininaya
I used a Pythagorean identity in both ways.
myininaya
  • myininaya
way 1 was the way I was trying to get you to go about it the first time around
myininaya
  • myininaya
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.

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