The statement given is:
\[ \cos(a+b) \cos(a-b) = \cos^2 a - \cos^2 b \]
To determine if this is an identity, we can use a well-known trigonometric identity that states:
\[ \cos(x) \cos(y) = \frac{1}{2} [\cos(x+y) + \cos(x-y)] \]
Using \( x = a + b \) and \( y = a - b \), we find:
\[ \cos(a+b) \cos(a-b) = \frac{1}{2} [\cos((a+b) + (a-b)) + \cos((a+b) - (a-b))] \]
This simplifies to:
\[ \cos(a+b) \cos(a-b) = \frac{1}{2} [\cos(2a) + \cos(2b)] \]
Now, let's rewrite the right side \(\cos^2 a - \cos^2 b\) using the difference of squares:
\[ \cos^2 a - \cos^2 b = (\cos a - \cos b)(\cos a + \cos b) \]
So, both sides of the equation appear to represent different forms, and it doesn't look like it's true all the time.
To check the validity of the identity, we can plug in specific values (e.g., \(a = 0\), \(b = 0\)) or general values, and upon calculation, we find that it does not hold for all values of \(a\) and \(b\).
Thus,
\[ \cos(a+b) \cos(a-b) \neq \cos^2 a - \cos^2 b \]
in general.
The answer is False.