Express (sin2x)/(1 - 2cos2x) in terms of tan x

[Monkeyseat]

Hi,

Question

Express (sin2x)/(1 - 2cos2x) in terms of tanx.

Working

I'm not sure what direction to go with this. I did:

(2sinxcosx)/(1-2(cos[sup:1l2jdbd2]2[/sup:1l2jdbd2]x - sin[sup:1l2jdbd2]2[/sup:1l2jdbd2]x))

I tried to get the numerator in terms of sinx and the denominator in terms of cosx, but I couldn't quite get it in that form. Alternatively, I thought that I might need to try and get it in the form tan2x and then use the double angle identity to get it in terms of tanx, but again, I wasn't sure how to get it that way.

The book says the answer is:

(2tanx)/(3tan[sup:1l2jdbd2]2[/sup:1l2jdbd2]x - 1)

Any help would be greatly appreciated.

Thanks.

 

[Deleted member 4993]

Hi,

Question

Express (sin2x)/(1 - 2cos2x) in terms of tanx.

Working

I'm not sure what direction to go with this. I did:

(2sinxcosx)/(1-2(cos[sup:2z556qp8]2[/sup:2z556qp8]x - sin[sup:2z556qp8]2[/sup:2z556qp8]x))

Divide the numerator and the denominator by cos[sup:2z556qp8]2[/sup:2z556qp8](x) - see where does it take you.

 

[Monkeyseat]

Thanks for the reply. I think I've managed to get a bit further, but I've hit another dead end...

If you could give me some more help it would be much appreciated. Thanks.

 

[Deleted member 4993]

Express (sin2x)/(1 - 2cos2x) in terms of tanx.

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

\(\displaystyle = \, \frac{2\sin(x)\cdot\cos(x)}{1-2\cos^2(x)+2\sin^2(x)}\)

\(\displaystyle = \, \frac{\frac{2\sin(x)\cdot\cos(x)}{\cos^2(x)}}{\frac{1-2\cos^2(x)+2\sin^2(x)}{cos^2(x)}}\)

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

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

and continue...