The simplification of frac{cos (90^{circ}- A | vedclass

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Question

(A) We use the trigonometric identities for complementary angles:$\cos (90^{\circ}- A ) = \sin A$$\sin (90^{\circ}- A ) = \cos A$$	an (90^{\circ}- A

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$\sin ^{2} A$

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(A) We use the trigonometric identities for complementary angles:$\cos (90^{\circ}- A ) = \sin A$$\sin (90^{\circ}- A ) = \cos A$$	an (90^{\circ}- A ) = \cot A$Substituting these into the expression:$\frac{\cos (90^{\circ}- A ) \sin (90^{\circ}- A )}{	an (90^{\circ}- A )} = \frac{\sin A \cdot \cos A}{\cot A}$Since $\cot A = \frac{\cos A}{\sin A}$,we have:$= \frac{\sin A \cdot \cos A}{\frac{\cos A}{\sin A}}$$= \sin A \cdot \cos A \cdot \frac{\sin A}{\cos A}$$= \sin A \cdot \sin A = \sin ^{2} A$

The simplification of $\frac{\cos (90^{\circ}- A ) \sin (90^{\circ}- A )}{\tan (90^{\circ}- A )}$ is .......

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