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$

Part $I$	Part $II$
$1.$ $\cos(90^\circ - \theta)$	$a.$ $\sec \theta$
$2.$ $\cot(90^\circ - \theta)$	$b.$ $\sin \theta$
$3.$ $\operatorname{cosec}(90^\circ - \theta)$	$c.$ $1$
	$d.$ $\tan \theta$

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

Similar Questions

Prove that,$\tan \theta + \tan (90^{\circ} - \theta) = \sec \theta \sec (90^{\circ} - \theta)$

Which of the following groups truly matches the data of Part $I$ with the data of Part $II$?

Part $I$ Part $II$

$1.$ $\cos(90^\circ - \theta)$ $a.$ $\sec \theta$

$2.$ $\cot(90^\circ - \theta)$ $b.$ $\sin \theta$

$3.$ $\operatorname{cosec}(90^\circ - \theta)$ $c.$ $1$

$d.$ $\tan \theta$

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If $\sin ^{2}(3 x+30^{\circ})+\cos ^{2}(2 x+45^{\circ})=1$,then $x = \dots$ (in $^{\circ}$)

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