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Measuring angles - class-XI
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$\dfrac{\cos(90-A)\sin(90-A)}{\tan (90-A)}$=
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A
$\sin^{2}A$
💡 Explanation:
$ = \dfrac{cos\,(90-A)\,sin\,(90-A)}{tan(90-A)}$
$ = \dfrac{sin\,A \times cos\,A}{\dfrac{cos\,A}{sin\,A}} = \dfrac{sin^{2}A\times cos\,A}{cos\,A}$
$ = sin^{2}A$