Question

Simplify cos2 Θ(1 + tan2 Θ)

Answer 1

I'm supposed to be using the Pythagorean Identity to do this...

Answer 2

@Abhisar

Answer 3

Does the '2' represent square?
Like \[\cos^2(1+\tan^2)\]

Answer 4

Yes, it does. Sorry about that!

Answer 5

You got the format correct the second time around.

Answer 6

Well, your first.

Answer 7

Do you know any trig pythagorean identities?

Answer 8

Ummmm, sin^2+cos^2=1

Answer 9

I think that's how it goes

Answer 10

With that, this problem can be solved!

Answer 11

You just have to change it around a bit

Answer 12

cos^2 Θ(1 + tan^2 Θ)
cos^2 Θ(1 +( sin^2 Θ/cos^2 Θ))
cos^2 Θ((cos^2 Θ + sin^2 Θ)/cos^2 Θ))
=1
source: http://www.mathskey.com/question2answer/

Answer 13

Look at this and see if you understand:\[\frac{ \sin^2+\cos^2 }{ \cos^2 }=\frac{ 1 }{ \cos^2 }\]

Answer 14

\[\frac{ \sin^2+\cos^2 }{ \cos^2 }=\frac{ \sin^2 }{ \cos^2 }+\frac{ \cos^2 }{ \cos^2 }=\]

Answer 15

Couldn't you simplify that to tan^2+cos=1/cos^2?

Answer 16

Exactly!

Answer 17

Now substitute that into the original equation to get...

Answer 18

Yes! I love it when shot's in the dark hit the target

Answer 19

Whoa, why are we substituting? I thought we were simplifying the same equation the whole time.

Answer 20

I'm sorry :/

Answer 21

Wait...

Answer 22

Ok...

Answer 23

I almost missed that!
\[\frac{ \sin^2 }{ \cos^2 }+\frac{ \cos^2 }{ \cos^2 }=\tan^2+1\]

Answer 24

which equals to 1/cos^2

Answer 25

Thats saying is that Tan^2=1

Answer 26

O crud, forgot a one

Answer 27

Nevermind

Answer 28

Nope.
\[\tan^2+1=\]

Answer 29

Isn't it saying 1=1?

Answer 30

\[\frac{ 1 }{ \cos^2 }\]

Answer 31

u can write 1+tan^2 (theta) = sec^2(theta)

Answer 32

Please do

Answer 33

sec^2(theta) will be write 1/cos^2(theta)

Answer 34

cos^2(theta)*(1/cos^2(theta)) =1

Answer 35

I'm hopelessly confused. I'm so sorry

Answer 36

Do we still work at \(cos^2 (1+ tan^2)\)??

Answer 37

Yes, we are

Answer 38

We're trying to simplify it using Pythagorean Identity/ies

Answer 39

ok, distribute cos^2 into the bracket, we have
\(cos^2 + cos^2* tan^2\) ok then?

Answer 40

now, tan =\(\dfrac{sin}{cos}\) so that \(tan^2 =\dfrac{sin^2}{cos^2}
\)

Answer 41

Why are there two cos^2 in your first reply?

Answer 42

replace to expression, you have
\(cos^2 + cos^2*\dfrac{sin^2}{cos^2}= cos^2 + sin^2 =1\)

Answer 43

I'm sorry, I'm taking an online course that never explained any of this, and math isn't my strong suit anyways.