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 $1$.

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

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