Answer
The equation is an identity.
Work Step by Step
Graphing utility used: online calculator, desmos.com.
(see below)
The graphs of the two expressions appear to be one and the same,
implying that the equation is an identity.
To verify,
recognize a difference of squares on the LHS.
$LHS=(\cos^{2}\theta+\sin^{2}\theta)(\cos^{2}\theta-\sin^{2}\theta)$
First parentheses: Pyth. identity:$\cos^{2}\theta+\sin^{2}\theta=1$
Second parentheses: Double angle $\cos^{2}\theta-\sin^{2}\theta=\cos 2\theta$
(First form)
$ LHS=\cos 2\theta$
The Second form of $\cos 2A$:
$\cos 2A=2\cos^{2}A-1$
$\mathrm{L}\mathrm{H}\mathrm{S}=2\cos^{2}\theta-1 =RHS$
The equation is an identity.
