$\dfrac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$
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?