The value of cos y cos (frac{pi}{2} - x) - co | vedclass

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Question

(D) Given expression: $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2

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$x = n\pi - \frac{\pi}{4} + y, (n \in I)$

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(D) Given expression: $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y) = 0$Using $\cos (\frac{\pi}{2} - 	heta) = \sin 	heta$ and $\sin (\frac{\pi}{2} - 	heta) = \cos 	heta$,the expression becomes:$\cos y \sin x - \sin y \cos x + \sin y \sin x + \cos x \cos y = 0$Rearranging the terms:$(\sin x \cos y - \cos x \sin y) + (\sin x \sin y + \cos x \cos y) = 0$Using the identities $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$\sin(x - y) + \cos(x - y) = 0$Dividing by $\cos(x - y)$ (assuming $\cos(x - y) 
eq 0$):$	an(x - y) = -1$$	an(x - y) = 	an(-\frac{\pi}{4})$$x - y = n\pi - \frac{\pi}{4}$$x = n\pi - \frac{\pi}{4} + y, (n \in I)$

The value of $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y)$ is zero,if

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