Trigo!

Since $\sin(\alpha) = 2\sin\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{2}\right)$ , if we divide both the numerator and denominator of the given expression by $\cos^{2}\left(\dfrac{\alpha}{2}\right)$ the expression becomes

$\dfrac{\sec^{2}\left(\dfrac{\alpha}{2}\right) - 2\tan^{2}\left(\dfrac{\alpha}{2}\right)}{\sec^{2}\left(\dfrac{\alpha}{2}\right) + 2\tan\left(\dfrac{\alpha}{2}\right)} = \dfrac{1 - \tan^{2}\left(\dfrac{\alpha}{2}\right)}{\tan^{2}\left(\dfrac{\alpha}{2}\right) + 2\tan\left(\dfrac{\alpha}{2}\right) + 1}$ ,

where the identity $\sec^{2}(x) = \tan^{2}(x) + 1$ was used. The expression can now be factored as

$\dfrac{\left(1 - \tan\left(\dfrac{\alpha}{2}\right)\right)\left(1 + \tan\left(\dfrac{\alpha}{2}\right)\right)}{\left(1 + \tan\left(\dfrac{\alpha}{2}\right)\right)^{2}} = \dfrac{1 - \tan\left(\dfrac{\alpha}{2}\right)}{1 + \tan\left(\dfrac{\alpha}{2}\right)} = \boxed{\dfrac{1 - k}{1 + k}}$ , assuming $\sin(\alpha) \ne -1$ .