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.
$LHS=\sec^{2}B-1$
Apply the Pyth. identity: $1 +\tan^{2}\theta=\sec^{2}\theta$
$LHS=1 +\tan^{2}B-1=\tan^{2}B=RHS$
The equation is an identity.