Prove that: $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$

solution:

It is known that

$\sin \mathrm{A}+\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right), \cos \mathrm{A}+\cos \mathrm{B}=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$

L.H.S. $=\frac{(\sin 7 \mathrm{x}+\sin 5 \mathrm{x})+(\sin 9 \mathrm{x}+\sin 3 \mathrm{x})}{(\cos 7 \mathrm{x}+\cos 5 \mathrm{x})+(\cos 9 \mathrm{x}+\cos 3 \mathrm{x})}$

$=\frac{\left[2 \sin \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \sin \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}{\left[2 \cos \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \cos \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}$

$=\frac{[2 \sin 6 x \cdot \cos x]+[2 \sin 6 x \cdot \cos 3 x]}{[2 \cos 6 x \cdot \cos x]+[2 \cos 6 x \cdot \cos 3 x]}$

$=\frac{2 \sin 6 x[\cos x+\cos 3 x]}{2 \cos 6 x[\cos x+\cos 3 x]}$

$=\tan 6 x$

$=$ R.H.S.