Verify the identity. $$ \cos (\alpha+\beta) \cos (\alpha-\beta)=\cos ^{2} \alpha-\sin ^{2} \beta $$ Write the left side of the identity using the sum and difference formula. ( $$ \cos \alpha \cos \beta-\sin \alpha \sin \beta)(\cos \alpha \cos \beta+\sin \alpha \sin \beta $$ ) (Type the terms of your expression in the same order as they appear in the original expression. Use the distributive property to simplify the expression from the previous step. $$ \cos ^{2} \alpha \cos ^{2} \beta-\sin ^{2} \alpha \sin ^{2} \beta $$ (Simplify your answer.) Substitute $$ 1-\sin ^{2} \beta $$ for $$ \cos ^{2} \beta $$ in the expression from the previous step and distribute. $$ \square $$

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After substituting $$ 1 - \sin^{2} \beta $$ for $$ \cos^{2} \beta $$ in the expression $$ \cos^{2} \alpha \cos^{2} \beta - \sin^{2} \alpha \sin^{2} \beta $$ and distributing, we get:

$$
\cos^{2} \alpha (1 - \sin^{2} \beta) - \sin^{2} \alpha \sin^{2} \beta = \cos^{2} \alpha - \cos^{2} \alpha \sin^{2} \beta - \sin^{2} \alpha \sin^{2} \beta
$$

Notice that $$ \cos^{2} \alpha + \sin^{2} \alpha = 1 $$. Therefore, we can factor out $$ \sin^{2} \beta $$:

$$
\cos^{2} \alpha - \sin^{2} \beta (\cos^{2} \alpha + \sin^{2} \alpha) = \cos^{2} \alpha - \sin^{2} \beta \cdot 1 = \cos^{2} \alpha - \sin^{2} \beta
$$

So, the simplified expression is:

$$
\boxed{\cos^{2} \alpha - \sin^{2} \beta}
$$
After substituting $$ 1 - \sin^{2} \beta $$ for $$ \cos^{2} \beta $$ and distributing, the expression simplifies to $$ \cos^{2} \alpha - \sin^{2} \beta $$.

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