Question

Can someone help with this? 1/cos^2a+1-3tan^2a=0

Answer 1

\[\frac{1}{\cos ^2a}+1-3\tan ^2a=0\]

Answer 2

Maybe that's it.

Answer 3

Yes your second one is the question I'm given.

Answer 4

\[\frac{1}{\cos ^2a}+\frac{\cos ^2a}{\cos ^2a}-\frac{3\sin ^2a}{\cos ^2a}=0\]

Answer 5

Does that help?

Answer 6

I would now add the fractions and replace sin^2 with 1-cos^2

Answer 7

How did you get the middle part?

Answer 8

Oh from +1

Answer 9

yes

Answer 10

so 1cos^2a/cos^2a-3sin^2a/cos^2a=0

Answer 11

Then 1cos^2a/cos^2a- 3(1-cos^2)/cos^2a=0?

Answer 12

Ok I think I can get it form here.. Thanks so much for the help!

Answer 13

yw

Answer 14

Make sure that is 1+cos^2a not 1cos^2a

Answer 15

\[\frac{1+\cos ^2a-3(1-\cos ^2a)}{\cos ^2a}=0\]

Answer 16

\[\frac{1+\cos ^2a-3+3\cos ^2a}{\cos ^2a}=0\]

Answer 17

\[\frac{4\cos ^2a-2}{\cos ^2a}=0\]

Answer 18

So you are saying that:
\[\frac{3(1-\cos ^2a)}{\cos ^2a}=0\]

Answer 19

That does not equal 0

Answer 20

\[3\tan ^2a=3\frac{\sin ^2a}{\cos ^2a}=\frac{3}{1}\frac{\sin ^2a}{\cos ^2a}=\frac{3(1-\cos ^2a)}{\cos ^2a}\]

Answer 21

They are the same thing.

Answer 22

no

Answer 23

\[\frac{1-\cos ^2a}{\cos ^2a}=\frac{1}{\cos ^2a}-\frac{\cos ^2a}{\cos ^2a}\]

Answer 24

You lost me. I thought we had a equation.

Answer 25

Yeah I think I need to go back a lot of steps, I must have made a mistake somewhere. :(

Answer 26

1+cos^2a/cos^2a-3(1-cos^2)/cos^2a=0\[\frac{ 1+\cos^2a }{ \cos^2a } -\frac{ 3(1-\cos^2a) }{ \cos^2a }=0\] I think thats what I'm up to.

Answer 27

Yes. I agree.

Answer 28

So do I add these fractions?

Answer 29

That's what I did. Or I guess you could put the one fraction on the right side and then cross multiply.

Answer 30

so would that be \[\frac{ 2(\sin a)^2+2 }{ \cos^2a }=0\]

Answer 31

Where did the sin come from? We had all cos

Answer 32

\[\frac{1+\cos ^2a-3(1-\cos ^2a)}{\cos ^2a}=0\]

Answer 33

\[\frac{1+\cos ^2a-3+3\cos ^2a}{\cos ^2a}=0\]

Answer 34

\[\frac{4\cos ^2a-2}{\cos ^2a}=0\]

Answer 35

If a fraction is 0, its numerator is 0. Therefore:
\[4\cos ^2a-2=0\]

Answer 36

\[4\cos ^2a=2\]
\[\cos ^2a=\frac{1}{2}\]
\[\cos a=\pm\frac{\sqrt{2}}{2}\]
\[a=\frac{\pi}{4}+\frac{\pi}{4}n\]

Answer 37

Ok that makes sense. I obviously need a lot more trigonometry practise though :/
Thanks sooooo much for the help you're awesome!

Answer 38

yw