Hello, Rolita!
If no one is responding, it's because it is not an identity.
I think this is what you meant: \(\displaystyle \;\cos(A\,-\,B)\cdot\cos(A\,+\,B)\;=\;\cos^2A\,-\,\sin^2B\)
The left side is: \(\displaystyle \:[\cos A\cdot\cos B\,+\,\sin A\cdot\sin B][\cos A\cdot\cos B\,-\,\sin A\cdot\sin B]\)
\(\displaystyle \;\;\;=\;\;\;\cos^2A\cdot\underbrace{\cos^2B}\;\;\;-\;\;\;\underbrace{\sin^2A}\cdot\sin^2B\)
\(\displaystyle \;\;\;=\;\cos^2A(\overbrace{1\,-\,\sin^2B})\,-\,(\overbrace{1\,-\cos^2A})\sin^2B\)
\(\displaystyle \;\;\;=\;\cos^2A\,-\,\cos^2A\cdot\sin^2B\,-\,\sin^2B \,+\,\cos^2A\cdot\sin^2B\)
\(\displaystyle \;\;\;=\;\cos^2A\,-\,\sin^2B\)
