Answer: see proof below
Step-by-step explanation:
Use the following Sum to Product Identities:
[tex]\cos A+\cos B=2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B}{2}\bigg)\\\\\\\sin A-\sin B=2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)[/tex]
Use Even/Odd Identities: cos (-A) = - cos A
sin (-A) = sin (A)
Use Quotient Identity: [tex]\cot A=\dfrac{\cos A}{\sin A}[/tex]
Proof LHS → RHS:
[tex]\text{LHS:}\qquad \qquad \qquad \dfrac{\cos x+\cos 7x}{\sin x - \sin 7x}[/tex]
[tex]\text{Sum to Product:}\qquad \dfrac{2\cos \bigg(\dfrac{x+7x}{2}\bigg)\cdot \cos \bigg(\dfrac{x-7x}{2}\bigg)}{2\cos \bigg(\dfrac{x+7x}{2}\bigg)\cdot \sin \bigg(\dfrac{x-7x}{2}\bigg)}[/tex]
[tex]\text{Simplify:}\qquad \qquad \quad \dfrac{2\cos \bigg(\dfrac{8x}{2}\bigg)\cdot \cos \bigg(\dfrac{-6x}{2}\bigg)}{2\cos \bigg(\dfrac{8x}{2}\bigg)\cdot \sin \bigg(\dfrac{-6x}{2}\bigg)}[/tex]
[tex]= \dfrac{ \cos (-3x)}{\sin(-3x)}[/tex]
[tex]\text{Even Odd:}\qquad \quad \dfrac{ -\cos (3x)}{\sin(3x)}[/tex]
[tex]\text{Quotient:}\qquad \quad -\cot(3x)[/tex]
LHS = RHS [tex]\checkmark[/tex]
