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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?

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Yes, it is.${}$ – David Mitra Jul 15 '12 at 1:38
Please try to write more specific titles – Dylan Moreland Jul 15 '12 at 2:06
up vote 1 down vote accepted

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.

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