The value of

(c) $\tan \alpha / 2$

$\frac{\sin 5 \alpha-\sin 3 \alpha}{\cos 5 \alpha+2 \cos 4 \alpha+\cos 3 \alpha}=\frac{\sin 5 \alpha-\sin 3 \alpha}{\cos 5 \alpha+\cos 3 \alpha+2 \cos 4 \alpha}$

$=\frac{2 \sin \alpha \cos 4 \alpha}{2 \cos 4 \alpha \cos \alpha+2 \cos 4 \alpha}$

$=\frac{2 \sin \alpha \cos 4 \alpha}{2 \cos 4 \alpha(\cos \alpha+1)}$

$=\frac{\sin \alpha}{\cos \alpha+1}$

$=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}}$

$=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^{2} \frac{\alpha}{2}}$

$=\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$

$=\tan \frac{\alpha}{2}$