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

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Question

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

Vedclass · Accepted Answer

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

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(B) Given expression: $\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}}$Using the identity $\cos 	heta = \sin(90^\circ - 	heta)$:$= \frac{{\sin (B + A) + \sin(90^\circ - (B - A))}}{{\sin (B - A) + \sin(90^\circ - (B + A))}}$$= \frac{{\sin (B + A) + \sin(90^\circ - B + A)}}{{\sin (B - A) + \sin(90^\circ - B - A)}}$Using the sum-to-product formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:Numerator: $2 \sin(\frac{B+A+90^\circ-B+A}{2}) \cos(\frac{B+A-90^\circ+B-A}{2}) = 2 \sin(45^\circ+A) \cos(B-45^\circ)$Denominator: $2 \sin(\frac{B-A+90^\circ-B-A}{2}) \cos(\frac{B-A-90^\circ+B+A}{2}) = 2 \sin(45^\circ-A) \cos(B-45^\circ)$Since $\cos(B-45^\circ) = \cos(45^\circ-B)$:$= \frac{2 \sin(A+45^\circ) \cos(45^\circ-B)}{2 \sin(45^\circ-A) \cos(45^\circ-B)} = \frac{\sin(A+45^\circ)}{\sin(45^\circ-A)}$Using $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$:$= \frac{\sin A \cos 45^\circ + \cos A \sin 45^\circ}{\sin 45^\circ \cos A - \cos 45^\circ \sin A}$Dividing numerator and denominator by $\sin 45^\circ$ (or multiplying by $\sqrt{2}$):$= \frac{\sin A + \cos A}{\cos A - \sin A} = \frac{\cos A + \sin A}{\cos A - \sin A}$.

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

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