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

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Question

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

Vedclass · Accepted Answer

$	an (A + B)$

Vedclass · Answer

(A) Given expression: $\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B}$Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:Numerator $= \sin(A+B) \sin(A-B)$Multiply numerator and denominator by $2$:$= \frac{2 \sin(A+B) \sin(A-B)}{2 \sin A \cos A - 2 \sin B \cos B}$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$:$= \frac{2 \sin(A+B) \sin(A-B)}{\sin 2A - \sin 2B}$Using the identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$:$\sin 2A - \sin 2B = 2 \cos(\frac{2A+2B}{2}) \sin(\frac{2A-2B}{2}) = 2 \cos(A+B) \sin(A-B)$Substituting back:$= \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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