# Trigonometric Identities

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

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?

-
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

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.

-