This has been alluding me for a week and I am now down to the wire
Prove that (sin^2(x)) / (1-cos(x)) equals (sec(x)+1)/(sec(x))
any help,hints,clues or outright answers would be appreciated!
Thanks,
mj
Consider the $\displaystyle LHS$
$\displaystyle \frac {\sin^2 x}{1 - \cos x} = \frac {1 - \cos^2 x}{1 - \cos x}$ ...........Pythagorean identity
$\displaystyle = \frac {(1 + \cos x )( 1 - \cos x)}{1 - \cos x}$ ..........Numerator was the difference of two squares
$\displaystyle = 1 + \cos x$ ..........................cancel
$\displaystyle = \frac {\sec x}{\sec x} + \frac 1{\sec x}$ .....................algebraic manipulation
$\displaystyle = \frac {\sec x + 1}{\sec x}$ ...........................combine the fractions
$\displaystyle = RHS$
Thus, $\displaystyle LHS \equiv RHS$, as desired
$\displaystyle \frac{\sin^2(x)}{1-\cos(x)}=\frac{\sec(x)+1}{\sec(x)}$
we know that $\displaystyle \sin^2(x)=1-\cos^2(x)$
and we also know that $\displaystyle a^2-b^2=(a+b)(a-b)$
so $\displaystyle \sin^2(x)=1-\cos^2(x)=(1+\cos(x))(1-\cos(x))$
So
now subbing we get
$\displaystyle \frac{(1+\cos(x))(1-\cos(x))}{1-\cos(x)}=\frac{\sec(x)+1}{\sec(x)}$
Cancelling the left side and splitting the left we get
$\displaystyle 1+\cos(x)=\frac{\sec(x)}{\sec(x)}+\frac{1}{\sec(x) }$
now we know that $\displaystyle \frac{\sec(x)}{\sec(x)}=1$
and $\displaystyle \sec(x)=\frac{1}{\cos(x)}$
so $\displaystyle \frac{1}{\sec(x)}=\frac{1}{\frac{1}{\cos(x)}}=\cos (x)$
imputting back in gives
$\displaystyle 1+\cos(x)=1+\cos(x)$