# Another proving identities problem

• Feb 11th 2010, 06:06 AM
Croton
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!
• Feb 11th 2010, 06:38 AM
dedust
Quote:

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!

$\frac{\ tan(x) - \ sin(x)}{\sin^3(x)} =$ $\frac{\sin x - \sin x \cos x}{\cos x \sin x \sin^2 x} = \frac{1 - \cos x}{\cos x \sin^2 x}$ $\times \frac{1 + \cos x}{1 + \cos x} = \frac{\sin^2 x}{\sin^2 x \cos x (1 + \cos x )} = \frac{1}{\cos x + \cos^2 x}$
• Feb 11th 2010, 06:39 AM
pencil09
Quote:

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:

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

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

Code:

remember

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

so,

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

and
cancel

$1-cos(x)$

• Feb 11th 2010, 06:43 AM
Calculus26
See attachment
• Feb 11th 2010, 07:22 AM
Croton
That helped a lot, thanks!