# Thread: Another proving identities problem

1. ## Another proving identities problem

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

Prove that:

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

Help appreciated as always!

2. Originally Posted by Croton
I've stumbled across another problem that I'm having a hard time solving.

Prove that:

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

Help appreciated as always!
$\frac{\ tan(x) - \ sin(x)}{\sin^3(x)} =$ $\frac{\sin x - \sin x \cos x}{\cos x \sin x \sin^2 x} = \frac{1 - \cos x}{\cos x \sin^2 x}$ $\times \frac{1 + \cos x}{1 + \cos x} = \frac{\sin^2 x}{\sin^2 x \cos x (1 + \cos x )} = \frac{1}{\cos x + \cos^2 x}$

3. Originally Posted by Croton
I've stumbled across another problem that I'm having a hard time solving.

Prove that:

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

Help appreciated as always!

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

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

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

cancel sin(x)

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

Code:


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$\sin^2(x)=1-\ cos(x)$

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

Code:
remember

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$(a^2-b^2)=(a-b)(a+b)$
so,

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$1-\cos^2(x)=(1-\cos(x))(1+\cos(x))$
and
cancel

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$1-cos(x)$

4. See attachment

5. That helped a lot, thanks!