Another Identity

**IDontunderstand** · December 9th 2010, 06:43 PM

I have gotten stuck again on another problem.

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

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So then I multiplied by the recip. and simplified and worked with the other fraction to get:\
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So I can do this
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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.

**Prove It** · December 9th 2010, 06:51 PM

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

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

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

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

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

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

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

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

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

**SammyS** · December 9th 2010, 07:36 PM

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:

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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.

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