I've stumbled across another problem that I'm having a hard time solving.

Prove that:

$\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}=\frac{1}{\cos(x) + \ cos^2(x)}$

Help appreciated as always!

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- Feb 11th 2010, 06:06 AMCrotonAnother proving identities problem
I've stumbled across another problem that I'm having a hard time solving.

Prove that:

$\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}=\frac{1}{\cos(x) + \ cos^2(x)}$

Help appreciated as always! - Feb 11th 2010, 06:38 AMdedust
$\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)} = $$\displaystyle \frac{\sin x - \sin x \cos x}{\cos x \sin x \sin^2 x} = \frac{1 - \cos x}{\cos x \sin^2 x}$$\displaystyle \times \frac{1 + \cos x}{1 + \cos x} = \frac{\sin^2 x}{\sin^2 x \cos x (1 + \cos x )} = \frac{1}{\cos x + \cos^2 x}$

- Feb 11th 2010, 06:39 AMpencil09

$\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}$

$\displaystyle \frac{ \frac{\ sin(x)}{\ cos(x)} - \ sin(x)}{\sin^3(x)}$

$\displaystyle \frac{\ sin(x)(\frac{1}{\ cos(x)} - 1)}{\sin^3(x)}$

cancel sin(x)

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

Code:`$\displaystyle \sin^2(x)=1-\ cos(x)$`

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

Code:`remember $\displaystyle (a^2-b^2)=(a-b)(a+b)$`

so, $\displaystyle 1-\cos^2(x)=(1-\cos(x))(1+\cos(x))$

and

cancel $\displaystyle 1-cos(x)$

- Feb 11th 2010, 06:43 AMCalculus26
See attachment

- Feb 11th 2010, 07:22 AMCroton
That helped a lot, thanks!