Answer: see proof below
Step-by-step explanation:
Use the following Sum to Product Identities:
[tex]\sin x-\sin y=2\cos\bigg(\dfrac{x+y}{2}\bigg)\sin\bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x-\cos y=2\cos\bigg(\dfrac{x+y}{2}\bigg)\cos\bigg(\dfrac{x-y}{2}\bigg)[/tex]
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \qquad \qquad \dfrac{\sin 2A+\sin 5A-\sin A}{\cos A+\cos 2A+\cos 5A}[/tex]
[tex]\text{Regroup:}\qquad \qquad \qquad \dfrac{\sin 2A+(\sin 5A-\sin A)}{\cos 2A+(\cos 5A+\cos A)}[/tex]
[tex]\text{Sum to Product:}\qquad \dfrac{\sin 2A+2\cos\bigg(\dfrac{5A+A}{2}\bigg)\sin\bigg(\dfrac{5A-A}{2}\bigg)}{\cos2A+\cos\bigg(\dfrac{5A+A}{2}\bigg)\cos\bigg(\dfrac{5A-A}{2}\bigg)}[/tex]
[tex]\text{Simplify:}\qquad \qquad \qquad \dfrac{\sin 2A+2\cos 3A\sin 2A}{\cos 2A+2\cos 3A\cos 2A}[/tex]
[tex]\text{Factor:}\qquad \qquad \qquad \dfrac{\sin 2A(1+2\cos 3A)}{\cos 2A(1+2\cos 3A)}[/tex]
[tex]\text{Simplify:}\qquad \qquad \qquad \dfrac{\sin 2A}{\cos 2A}\\\\.\qquad \qquad \qquad \qquad =\tan 2A[/tex]
LHS = RHS: tan 2A = tan 2A [tex]\checkmark[/tex]
