The expression frac{tan left( frac{3pi}{2} - | vedclass

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Question

(A) Using trigonometric identities:$	an \left( \frac{3\pi}{2} - \alpha ight) = \cot \alpha$$\cos \left( \frac{3\pi}{2} - \alpha ight) = -\sin \al

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(A) Using trigonometric identities:$	an \left( \frac{3\pi}{2} - \alpha ight) = \cot \alpha$$\cos \left( \frac{3\pi}{2} - \alpha ight) = -\sin \alpha$$\cos (2\pi - \alpha ) = \cos \alpha$$\cos \left( \alpha - \frac{\pi}{2} ight) = \sin \alpha$$\sin (\pi - \alpha ) = \sin \alpha$$\cos (\pi + \alpha ) = -\cos \alpha$$\sin \left( \alpha - \frac{\pi}{2} ight) = -\cos \alpha$Substituting these into the expression:$= \frac{(\cot \alpha )(-\sin \alpha )}{\cos \alpha } + (\sin \alpha )(\sin \alpha ) + (-\cos \alpha )(-\cos \alpha )$$= \frac{(\frac{\cos \alpha }{\sin \alpha })(-\sin \alpha )}{\cos \alpha } + \sin^2 \alpha + \cos^2 \alpha$$= -1 + 1 = 0$

The expression $\frac{\tan \left( \frac{3\pi}{2} - \alpha \right) \cos \left( \frac{3\pi}{2} - \alpha \right)}{\cos (2\pi - \alpha )} + \cos \left( \alpha - \frac{\pi}{2} \right) \sin (\pi - \alpha ) + \cos (\pi + \alpha ) \sin \left( \alpha - \frac{\pi}{2} \right)$ when simplified reduces to:

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