1. ## Another Identity

I have gotten stuck again on another problem.

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

So this is what I did was made one fraction in the denominator:

$\displaystyle \frac {2 tan x}{(cos^2x - sin^2x)/cos^2x}+\frac{1}{2 cos^2x-1}=\frac{cos x + sin x}{cos x - sin x}$

So then I multiplied by the recip. and simplified and worked with the other fraction to get:\
$\displaystyle \frac {2 sinx cosx }{cos^2x - sin^2x}+\frac{1}{cos^2x - sin^2x}=\frac{cos x + sin x}{cos x - sin x}$

So I can do this
$\displaystyle \frac {2 sin x cos x +1 }{(cos x + sin x)(cos x - sin x)}=\frac{cos x + sin x}{cos x - sin x}$

I do not see how I could get the top to factor out the top and cancel to get this fraction. I have a strong feeling that I did something wrong but do not know where or how to fix this.

Thank you.

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

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

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

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

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

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

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

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

$\displaystyle \displaystyle = \frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}$.

3. ## Trig ident.

Originally Posted by IDontunderstand
I have gotten stuck again on another problem.

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

So this is what I did was made one fraction in the denominator:

$\displaystyle \frac {2 tan x}{(cos^2x - sin^2x)/cos^2x}+\frac{1}{2 cos^2x-1}=\frac{cos x + sin x}{cos x - sin x}$

Notice that the left hand side is:

$\displaystyle \displaystyle \tan 2x +\frac{1}{\cos (2x)}$

I think the right hand side might be more fruitful to work on.

$\displaystyle \displaystyle \frac{\cos x + \sin x}{\cos x - \sin x}$

Multiply top & bottom by $\displaystyle \cos x + \sin x$.

$\displaystyle \displaystyle \frac{(\cos x + \sin x)^2}{\cos^2 x - \sin^2 x}$

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

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

That should help.