# Another Identity

• December 9th 2010, 07:43 PM
IDontunderstand
Another Identity
I have gotten stuck again on another problem.

[LaTeX ERROR: Convert failed]

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

[LaTeX ERROR: Convert failed]

So then I multiplied by the recip. and simplified and worked with the other fraction to get:\
[LaTeX ERROR: Convert failed]

So I can do this
[LaTeX ERROR: Convert failed]

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.
• December 9th 2010, 07:51 PM
Prove It
$\displaystyle \frac{2\tan{x}}{1 - \tan^2{x}} + \frac{1}{2\cos^2{x} - 1} = \frac{\frac{2\sin{x}}{\cos{x}}}{1 - \frac{\sin^2{x}}{\cos^2{x}}} + \frac{1}{2\cos^2{x} - 1}$

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

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

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

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

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

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

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

$\displaystyle = \frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}$.
• December 9th 2010, 08:36 PM
SammyS
Trig ident.
Quote:

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

[LaTeX ERROR: Convert failed]

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

[LaTeX ERROR: Convert failed]

Notice that the left hand side is:

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

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

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

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

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

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

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

That should help.