The expression cos^2(A - B) + cos^2 B - 2cos( | vedclass

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(C) Let the expression be $E = \cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$.Using the identity $2\cos A\cos B = \cos(A - B) + \cos(A + B)$,we

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Dependent on $A$

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(C) Let the expression be $E = \cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$.Using the identity $2\cos A\cos B = \cos(A - B) + \cos(A + B)$,we get:$E = \cos^2(A - B) + \cos^2 B - \cos(A - B)[\cos(A - B) + \cos(A + B)]$$E = \cos^2(A - B) + \cos^2 B - \cos^2(A - B) - \cos(A - B)\cos(A + B)$$E = \cos^2 B - \cos(A - B)\cos(A + B)$Using the identity $\cos(A - B)\cos(A + B) = \cos^2 A - \sin^2 B$,we get:$E = \cos^2 B - (\cos^2 A - \sin^2 B)$$E = \cos^2 B + \sin^2 B - \cos^2 A$Since $\cos^2 B + \sin^2 B = 1$,we have:$E = 1 - \cos^2 A = \sin^2 A$.Since the result $\sin^2 A$ depends only on $A$,the expression is dependent on $A$.

The expression $\cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$ is

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