frac{sin^2 A - sin^2 B}{sin A cos A - sin B c | vedclass

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Question

(B) Given expression: $\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B}$Using the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$,th

Vedclass · Accepted Answer

$	an(A + B)$

Vedclass · Answer

(B) Given expression: $\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B}$Using the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$,the numerator becomes $2 \sin(A + B) \sin(A - B)$ if we multiply by $2/2$.Alternatively,multiply numerator and denominator by $2$:$= \frac{2(\sin^2 A - \sin^2 B)}{2 \sin A \cos A - 2 \sin B \cos B}$Using $2 \sin^2 	heta = 1 - \cos 2	heta$ and $2 \sin 	heta \cos 	heta = \sin 2	heta$:$= \frac{(1 - \cos 2A) - (1 - \cos 2B)}{\sin 2A - \sin 2B} = \frac{\cos 2B - \cos 2A}{\sin 2A - \sin 2B}$Using sum-to-product formulas $\cos C - \cos D = 2 \sin(\frac{C+D}{2}) \sin(\frac{D-C}{2})$ and $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$:$= \frac{2 \sin(A + B) \sin(A - B)}{2 \cos(A + B) \sin(A - B)}$$= \frac{\sin(A + B)}{\cos(A + B)} = 	an(A + B)$.

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

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