Answer: see proof below
Step-by-step explanation:
Use the following Double Angle Identities: cos 2A = 1 - 2sin²A
sin 2A = 2sinA · cosA
[tex]\text{Given:}\quad \tan A = \dfrac{x}{y}[/tex]
Proof LHS = RHS
y cos 2A + x sin 2A = y
x sin 2A = y - y cos 2A
x sin 2A = y(1 - cos 2A)
[tex]\dfrac{x}{y}=\dfrac{1-\cos2A}{sin\ 2A}[/tex]
[tex]\dfrac{x}{y}=\dfrac{1-(1-2sin^2A)}{2\sin A\cdot \cos A}[/tex]
[tex]\dfrac{x}{y}=\dfrac{2sin^2A}{2\sin A\cdot \cos A}[/tex]
[tex]\dfrac{x}{y}=\dfrac{\sin A}{\cos A}[/tex]
[tex]\dfrac{x}{y}=\tan A}[/tex]
This is a TRUE statement since it was given that [tex]\tan A = \dfrac{x}{y}[/tex]
