Answer
The identity is verified.
$cos(x+y)~cos(x-y)=cos^2x-sin^2y$
Work Step by Step
$cos(u+v)=cos~u~cos~v-sin~u~sin~v$
$cos(u-v)=cos~u~cos~v+sin~u~sin~v$
$cos^2x+sin^2x=1$
$cos^2x=1-sin^2x$
$sin^2x=1-cos^2x$
$cos(x+y)~cos(x-y)=(cos~x~cos~y-sin~x~sin~y)(cos~x~cos~y+sin~x~sin~y)=cos^2~x~cos^2~y-sin^2~x~sin^2~y=cos^2x(1-sin^2y)-(1-cos^2x)sin^2y=cos^2x-cos^2x~sin^2y-sin^2y+cos^2xsin^2y=cos^2x-sin^2y$
