how do i prove : tan^2 θ cos^2 θ + cos^2 θ = 1

- anonymous

how do i prove : tan^2 θ cos^2 θ + cos^2 θ = 1

- jamiebookeater

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- 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

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

- anonymous

I'm pretty confused

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## More answers

- myininaya

on how to factor?

- myininaya

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

- myininaya

how* (not hot)

- anonymous

no

- 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

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

yes

- myininaya

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

- myininaya

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

- 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

yes

- 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

cos^2?

- myininaya

right
so what does that leave us with?

- myininaya

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

- anonymous

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

- myininaya

And that equals?

- myininaya

you need to recall some trig identities for these trigonometric proofs

- myininaya

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

- anonymous

okay si... 1?

- myininaya

I will give you a hint:
Pythagorean Identity

- myininaya

yep

- anonymous

*so

- anonymous

okay i get it now thanks soooo much

- 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

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

- myininaya

I used a Pythagorean identity in both ways.

- myininaya

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

- 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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