sin (45^{circ}+theta)-cos (45^{circ}-theta) i | vedclass

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(B) We know that $\sin (90^{\circ}-	heta) = \cos 	heta$ and $\cos (90^{\circ}-	heta) = \sin 	heta$.Given expression: $\sin (45^{\circ}+	heta)-\co

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(B) We know that $\sin (90^{\circ}-	heta) = \cos 	heta$ and $\cos (90^{\circ}-	heta) = \sin 	heta$.Given expression: $\sin (45^{\circ}+	heta)-\cos (45^{\circ}-	heta)$Using the identity $\sin A = \cos (90^{\circ}-A)$:$\sin (45^{\circ}+	heta) = \cos (90^{\circ}-(45^{\circ}+	heta))$$= \cos (90^{\circ}-45^{\circ}-	heta)$$= \cos (45^{\circ}-	heta)$Substituting this back into the original expression:$\cos (45^{\circ}-	heta)-\cos (45^{\circ}-	heta) = 0$

$\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)$ is equal to

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