Question

Prove that:

(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x.

Answer 1

Solution:

Formula used:

$\sf{1).\;\cos A + \cos B = 2\cos\bigg(\dfrac{A + B}{2}\bigg)\;\cos\bigg(\dfrac{A-B}{2}\bigg)}$

$\sf{2).\;\sin A + \sin B = 2\cos\bigg(\dfrac{A + B}{2}\bigg)\;\sin\bigg(\dfrac{A-B}{2}\bigg)}$

So,

$\sf{\implies \dfrac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}=\cot 3x}$

We take LHS part,

$\sf{\implies \dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x + \sin 3x + \sin 2x}}$

$\sf{\implies \dfrac{(\cos 4x + \cos 2x)+\cos 3x}{(\sin 4x + \sin 2x)+\sin 3x}}$

$\sf{\implies \dfrac{\Bigg[2\cos \bigg(\dfrac{4x+2x}{2}\bigg)\;\cos\bigg(\dfrac{4x-2x}{2}\bigg)\Bigg]+\cos 3x}{\Bigg[ 2\sin \bigg(\dfrac{4x+2x}{2}\bigg) \;\cos\bigg(\dfrac{4x-2x}{2}\bigg)\Bigg]+\sin 3x}}$

$\sf{\implies \dfrac{2\cos 3x \cos x + \cos 3x}{2 \sin 3x \cos x + \sin 3x}}$

$\sf{\implies \dfrac{\cos 3x \;(2\cos x + 1)}{\sin 3x \;(2 \cos x + 1)}}$

${\boxed{\boxed{\sf{\implies \cot 3x}}}}$

LHS = RHS

Hence Proved!!