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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

$\sin ^{2} A$

Vedclass · Answer

(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

If $\operatorname{cosec} A = \sqrt{10}$,then $\sin A = \dots$

$\tan (90^\circ - \theta) = \ldots \ldots \ldots$

Given that $\sin \theta = \frac{a}{b},$ then $\cos \theta$ is equal to

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

Part $I$ Part $II$

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

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

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

$d.$ $\tan \theta$

Write 'True' or 'False' and justify your answer.
$\cos \theta = \frac{a^{2} + b^{2}}{2ab}$,where $a$ and $b$ are two distinct numbers such that $ab > 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module