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
equal,
implying that the equation is an identity.
To verify,
the common denominator on the LHS is
$(1-\sin x)(1+\sin x)=1-\sin^{2}x$
(difference of squares)
$LHS=\displaystyle \frac{1+\sin x+(1-\sin x)}{(1-\sin^{2}x)}$
... in the denominator, $1=\cos^{2}x+\sin^{2}x$
$=\displaystyle \frac{2}{\cos^{2}x+\sin^{2}x-\sin^{2}x}$
$=\displaystyle \frac{2}{\cos^{2}x}=2\sec^{2}x=RHS$
The equation is an identity.