can someone help in prove the identity:

is equal to2sin2θ1+cos2tan2θ

in english 2sin sqaured theta divided by 1 + cos2 eqauls to tan squared theta

any help by great

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- Dec 2nd 2010, 07:23 AMmaths99prove identity help!
can someone help in prove the identity:

is equal to__2sin____2__**θ****1+cos2****tan**2**θ**

**in english 2sin sqaured theta divided by 1 + cos2 eqauls to tan squared theta**

**any help by great** - Dec 2nd 2010, 08:56 AMelemental
Do you mean:

$\displaystyle {tan}^2 x = \frac{2{sin}^2 x}{1+cos 2x}$ ?

If so, then here's a proof:

$\displaystyle {tan}^2 x = \frac{1 - cos 2x}{1 + cos 2x}$

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

Substitute the new numerator.

$\displaystyle {tan}^2 x = \frac{2{sin}^2 x}{1 + cos 2x}$

QED. - Dec 2nd 2010, 09:31 AMArchie Meade
The left-hand-side is expressed in terms of sine and cosine.

We can also express the right-hand side in terms of sine and cosine...

$\displaystyle \displaystyle\frac{2sin^2\theta}{1+cos2\theta}=\le ft(\frac{sin\theta}{cos\theta}\right)^2=\frac{sin^ 2\theta}{cos^2\theta}$ ?

$\displaystyle sin^2\theta$ is common to both sides, so divide both sides by $\displaystyle sin^2\theta$

$\displaystyle \displaystyle\frac{2}{1+cos2\theta}=\frac{1}{cos^2 \theta}$ ?

Divide both sides by 2.

$\displaystyle \displaystyle\frac{1}{1+cos2\theta}=\frac{1}{2cos^ 2\theta}$ ?

Now you only need determine if the denominators are equal... - Dec 2nd 2010, 09:42 AMVonNemo19
Why do people just go and solve the problem for the newbie? This is his first post and who knows...Maybe this problem is part of a quiz for one of his online assignments....

Don't be too eager to please guys. That's all I'm saying.

Peace - Dec 2nd 2010, 09:44 AMmaths99
thanks for the help

- Dec 2nd 2010, 09:44 AMArchie Meade
- Dec 2nd 2010, 09:54 AMVonNemo19
Archie Meade. I've watched you for a while and you are genuinely here as a helper.

**Your reply**worked as a guide for the newb. So, rest assured, that I wasn't directing my post to you. Let's have a logic session, shall we?

$\displaystyle \text{ArchieMeade}\wedge\text{elemental}$

$\displaystyle \neg\text{ArchieMeade}$

$\displaystyle \rightarrow{elemental}$ - Dec 2nd 2010, 10:00 AMmaths99
just to let you know it wasnt part of an assignment or on line test. just im doing lots to gain practise in my civil engineer course, and was stuck on this. thanks for your help peace