If cos 9 alpha = sin alpha and 9 alpha

Question

(C) Given,$\cos 9 \alpha = \sin \alpha$ and $9 \alpha < 90^{\circ},$ which implies $9 \alpha$ is an acute angle.Using the trigonometric identity $\cos

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$1$

Vedclass · Answer

(C) Given,$\cos 9 \alpha = \sin \alpha$ and $9 \alpha Using the trigonometric identity $\cos A = \sin(90^{\circ} - A),$ we can rewrite the equation as:$\sin(90^{\circ} - 9 \alpha) = \sin \alpha$Since the sine functions are equal,their angles must be equal:$90^{\circ} - 9 \alpha = \alpha$Rearranging the terms to solve for $\alpha$:$10 \alpha = 90^{\circ}$$\alpha = 9^{\circ}$Now,we need to find the value of $	an 5 \alpha$:$	an 5 \alpha = 	an(5 	imes 9^{\circ}) = 	an 45^{\circ}$Since $	an 45^{\circ} = 1,$ the value is $1$.

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

If $\cos 9 \alpha = \sin \alpha$ and $9 \alpha < 90^{\circ},$ then the value of $\tan 5 \alpha$ is

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

Part $I$ Part $II$

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

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

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

$d.$ $\tan \theta$

Prove that:
If $\tan A = \frac{3}{4},$ then $\sin A \cos A = \frac{12}{25}$

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