Помогите пож 456,457.

Решите задачу:

$\frac{\sin2 \alpha }{2\cos \alpha } = \frac{2\sin \alpha\cos \alpha }{2\cos \alpha } =\sin \alpha \\\ \frac{\sin2 \alpha }{\sin \alpha } = \frac{2\sin \alpha\cos \alpha }{\sin \alpha } =2\cos \alpha$
$\frac{2\cos^2 \alpha }{\sin2 \alpha } = \frac{2\cos^2 \alpha }{2\sin \alpha\cos \alpha } = \frac{\cos \alpha }{\sin \alpha } =\mathop{\mathrm{ctg}} \alpha \\\ \frac{\cos2 \alpha }{\sin \alpha +\cos \alpha } = \frac{\cos^2 \alpha -\sin^2 \alpha }{\sin \alpha +\cos \alpha } = \frac{(\cos \alpha -\sin \alpha)(\cos \alpha +\sin \alpha) }{\sin \alpha +\cos \alpha } = \cos \alpha -\sin \alpha$

$\cos2 \alpha +\sin^2 \alpha =\cos^2 \alpha -\sin^2 \alpha+\sin^2 \alpha =\cos^2 \alpha \\\ \cos^2 \alpha -\cos2 \alpha =\cos^2 \alpha -\cos^2 \alpha+\sin^2 \alpha =\sin^2 \alpha$
$\frac{\mathrm{tg} \alpha }{1-\mathrm{tg}^2 \alpha } = \cfrac{ \frac{\sin \alpha }{\cos \alpha } }{1- \frac{\sin^2 \alpha}{\cos^2 \alpha} } = \cfrac{ \frac{\sin \alpha }{\cos \alpha } }{ \frac{\cos^2 \alpha -\sin^2 \alpha}{\cos^2 \alpha} } = \frac{\sin \alpha\cos^2 \alpha }{\cos \alpha(\cos^2 \alpha -\sin^2 \alpha) } = \\\ =\frac{\sin \alpha\cos \alpha }{\cos^2 \alpha -\sin^2 \alpha }=\frac{1}{2}\cdot\frac{\sin2 \alpha }{\cos2 \alpha } = \frac{\mathrm{tg} 2\alpha }{2}$
$\frac{\cos2 \alpha }{\sin \alpha +\cos \alpha } -\cos \alpha = \frac{\cos^2 \alpha-\sin^2 \alpha }{\sin \alpha +\cos \alpha } -\cos \alpha = \\\ = \frac{(\cos \alpha-\sin \alpha)(\cos \alpha+\sin \alpha) }{\sin \alpha +\cos \alpha } -\cos \alpha = \cos \alpha-\sin \alpha -\cos \alpha =-\sin \alpha$