frac{sin(B + A) + cos(B - A)}{sin(B - A) + co | vedclass

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Question

(B) Given expression: $E = \frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)}$Using the identities $\cos 	heta = \sin(90^\circ - 	heta)$:$E

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$\frac{\cos A + \sin A}{\cos A - \sin A}$

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(B) Given expression: $E = \frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)}$Using the identities $\cos 	heta = \sin(90^\circ - 	heta)$:$E = \frac{\sin(B + A) + \sin(90^\circ - (B - A))}{\sin(B - A) + \sin(90^\circ - (B + A))}$$E = \frac{\sin(B + A) + \sin(90^\circ - B + A)}{\sin(B - A) + \sin(90^\circ - B - A)}$Using $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:Numerator: $\sin(B+A) + \sin(90^\circ - B + A) = 2 \sin(\frac{B+A+90^\circ-B+A}{2}) \cos(\frac{B+A-90^\circ+B-A}{2}) = 2 \sin(45^\circ + A) \cos(B - 45^\circ)$Denominator: $\sin(B-A) + \sin(90^\circ - B - A) = 2 \sin(\frac{B-A+90^\circ-B-A}{2}) \cos(\frac{B-A-90^\circ+B+A}{2}) = 2 \sin(45^\circ - A) \cos(B - 45^\circ)$$E = \frac{2 \sin(45^\circ + A) \cos(B - 45^\circ)}{2 \sin(45^\circ - A) \cos(B - 45^\circ)} = \frac{\sin(45^\circ + A)}{\sin(45^\circ - A)}$Using $\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$:$E = \frac{\sin 45^\circ \cos A + \cos 45^\circ \sin A}{\sin 45^\circ \cos A - \cos 45^\circ \sin A} = \frac{\frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A}{\frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A} = \frac{\cos A + \sin A}{\cos A - \sin A}$.

$\frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)} = $

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