Trigonometric Identities: $\frac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$ $\dfrac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$  
Right Side:
$1-\cos\theta$ either stays the same, or can be $1-\dfrac{1}{\sec\theta}$  
Left Side:
$$\begin{align*}  
&= \dfrac{\sin^2\theta}{1+\cos\theta}\\  
&= \dfrac{1-\cos^2\theta}{1+\cos\theta}
&= \dfrac{(1-\cos\theta)(1+\cos\theta)}{1+cos\theta}
&= 1-\cos\theta
\end{align*}$$  
Is this correct?

Perhaps slightly simpler and shorter (FYI, what you did is correct):
$$\frac{\sin^2x}{1+\cos x}=1-\cos x\Longleftrightarrow \sin^2x=(1-\cos x)(1+\cos x)\Longleftrightarrow \sin^2x=1-\cos^2x$$
And since the last equality is just the trigonometric Pytahgoras Theorem we're done.
