# Prove Trig Identity

• Nov 11th 2008, 06:24 PM
Raj
Prove Trig Identity
$\displaystyle \frac{cos2x}{sin2x+1} = \frac{1-tanx}{1+tanx}$

Can't seem to prove this one, help please
• Nov 11th 2008, 06:58 PM
mr fantastic
Quote:

Originally Posted by Raj
$\displaystyle \frac{cos2x}{sin2x+1} = \frac{1-tanx}{1+tanx}$

Can't seem to prove this one, help please

Start by choosing a side.

I choose the right hand side.

Substitute $\displaystyle \tan x = \frac{\sin x}{\cos x}$. Multiply numerator and denominator by $\displaystyle \cos x$.

Now multiply numerator and denominator by $\displaystyle \cos x + \sin x$ and expand.

You should recognise the numerator as $\displaystyle \cos (2x)$. Simplify the denominator to recognise $\displaystyle \sin (2x)$.
• Nov 12th 2008, 07:30 AM
Raj
Where is the cosx coming from
• Nov 12th 2008, 11:52 AM
mr fantastic
Quote:

Originally Posted by Raj
Where is the cosx coming from

Did you do the substitution?

If you did it should be perfectly clear where the cos x comes from. You multiply top and bottom by cos x to get rid of the fraction.

You have to actually do each of the suggested steps before you'll see why you do them.