Help with trig identity

[lilygirl]

2 csc 2x=(1+tan^2x)/tan x


Ok So I have take the RHS and changed them to cos and sines and then used the half identities but I think I screwed up doing that and am not sure why.

Then I have also turned the RHS into sec^2x/tanx but thats not coming out correctly either.

Could anyone help advise me on what I am doing wrong?

 

[Deleted member 4993]

2 csc 2x=(1+tan^2x)/tan x


Ok So I have take the RHS and changed them to cos and sines and then used the half identities but I think I screwed up doing that and am not sure why.

Then I have also turned the RHS into sec^2x/tanx but thats not coming out correctly either.

Could anyone help advise me on what I am doing wrong?

Please show us your work - so that we understand the problem better.

Is the problem:

\(\displaystyle 2\cdot csc (2x) \, = \, \frac{1 \, + \, \tan^2(x)}{\tan(x)}\)

or something else....

 

[lilygirl]

Sorry, Im still trying to figure out how to put math problems on the computer.

The equation you wrote is the one I am looking for.

I did:

1+tan^2x =
tanx
sec^2x=
tanx
1
(cos^2x)tanx

 

[stapel]

Im still trying to figure out how to put math problems on the computer.

A good start might be to read the "Read Before Posting" thread, and follow one or both of the links provided for formatting information.

Thank you.

Eliz.

 

[Deleted member 4993]

Sorry, Im still trying to figure out how to put math problems on the computer.

The equation you wrote is the one I am looking for.

I did:

1+tan^2x =
tanx
sec^2x=
tanx
1
(cos^2x)tanx

\(\displaystyle \frac{sec^2(x)}{tan(x)} \,\)

\(\displaystyle = \frac{\frac{1}{\cos^2(x)}}{\frac{\sin(x)}{\cos(x)}}\)

\(\displaystyle = \frac{\frac{1}{\cos(x)}}{\sin(x)}\)

\(\displaystyle = \frac{1}{\cos(x)\cdot \sin(x)}\)

\(\displaystyle = \frac{2}{2\cdot\cos(x)\cdot \sin(x)}\)

Now continue....

 

[lilygirl]

Thank you! So I was actually doing it correctly? I just doubt myself sometimes. But thank you again!