# Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$

Question.

Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$

solution:

L.H.S. $=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$

$=\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x}$

$=\frac{2 \cos \left(\frac{4 \mathrm{x}+2 \mathrm{x}}{2}\right) \cos \left(\frac{4 \mathrm{x}-2 \mathrm{x}}{2}\right)+\cos 3 \mathrm{x}}{2 \sin \left(\frac{4 \mathrm{x}+2 \mathrm{x}}{2}\right) \cos \left(\frac{4 \mathrm{x}-2 \mathrm{x}}{2}\right)+\sin 3 \mathrm{x}}$

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

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

\begin{aligned} & \sin 3 x(2 \cos x+1) \\=& \cot 3 x=\text { R.H.S. } \end{aligned}