$\dfrac{\sin 4a}{1+\cos 4a}=\dfrac{2\sin 2a\cos 2a}{2\cos^{2}2a}=\dfrac{\sin 2a}{\cos 2a}$
$\dfrac{\sin 4a}{1+\cos 4a}=\dfrac{2\sin 2a\cos 2a}{2\cos^{2}2a}=\dfrac{\sin 2a}{\cos 2a}$$=\dfrac{2\sin a\cos a}{\cos^{2}a-\sin^{2}a}=\dfrac{\dfrac{2\sin a}{\cos a}}{1-\dfrac{\sin^{2}a}{\cos^{2}a}}$$=\dfrac{2\tan a}{1-\tan^{2}a}$
$\dfrac{\sin 4a}{1+\cos 4a}=\dfrac{2\sin 2a\cos 2a}{2\cos^{2}2a}=\dfrac{\sin 2a}{\cos 2a}$