If A+B=C,then the value of cos^2 A + cos^2 B | vedclass

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(A) Given that $A+B=C$. We need to evaluate the expression $E = \cos^2 A + \cos^2 B + \cos^2 C - 2 \cos A \cos B \cos C$. Using the identity $\cos^2 \

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(A) Given that $A+B=C$. We need to evaluate the expression $E = \cos^2 A + \cos^2 B + \cos^2 C - 2 \cos A \cos B \cos C$. Using the identity $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we have: $E = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} + \cos^2 C - 2 \cos A \cos B \cos C$ $E = 1 + \frac{1}{2}(\cos 2A + \cos 2B) + \cos^2 C - 2 \cos A \cos B \cos C$ Using $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$: $E = 1 + \cos(A+B) \cos(A-B) + \cos^2 C - 2 \cos A \cos B \cos C$ Since $A+B=C$,$\cos(A+B) = \cos C$: $E = 1 + \cos C \cos(A-B) + \cos^2 C - 2 \cos A \cos B \cos C$ $E = 1 + \cos C [\cos(A-B) + \cos C] - 2 \cos A \cos B \cos C$ Since $C = A+B$,$\cos C = \cos(A+B) = \cos A \cos B - \sin A \sin B$: $E = 1 + \cos C [\cos(A-B) + \cos(A+B)] - 2 \cos A \cos B \cos C$ Using $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$: $E = 1 + \cos C [2 \cos A \cos B] - 2 \cos A \cos B \cos C$ $E = 1 + 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C = 1$.

If $A+B=C$,then the value of $\cos^2 A + \cos^2 B + \cos^2 C - 2 \cos A \cos B \cos C$ is equal to

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