Simplify: \(\dfrac{\sin A \cdot \cos(90^\circ - A)}{\cot(90^\circ + A) \cdot \cos(180^\circ + A)}\)
\(\sin A\)
Reduce each shifted ratio using standard allied-angle identities.
For the numerator, \(\cos(90^\circ - A) = \sin A\), so the numerator becomes \(\sin A \cdot \sin A = \sin^2 A\).
In the denominator, \(\cot(90^\circ + A) = -\tan A\) and \(\cos(180^\circ + A) = -\cos A\), so their product is \((-\tan A)(-\cos A) = \tan A \cdot \cos A\).
Since \(\tan A \cdot \cos A = \frac{\sin A}{\cos A} \cdot \cos A = \sin A\), the denominator simplifies to \(\sin A\).
The whole expression is now \(\frac{\sin^2 A}{\sin A} = \sin A\).
Hence, the simplified value is \(\sin A\).
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