Originally Posted by
Croton 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!
$\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}$
$\displaystyle \frac{ \frac{\ sin(x)}{\ cos(x)} - \ sin(x)}{\sin^3(x)}$
$\displaystyle \frac{\ sin(x)(\frac{1}{\ cos(x)} - 1)}{\sin^3(x)}$
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 $\displaystyle 1-cos(x)$