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

Vedclass · Accepted Answer

$1$

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

Similar Questions

$\cos \theta = \frac{b}{\sqrt{a^2 + b^2}}$; where,$0 < \theta < 90^\circ$; then $\sin \theta = \dots$

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

If $\sin \theta = \frac{1}{2}$,then $\theta = \ldots \ldots \ldots \ldots$ (in $^\circ$)

If $\triangle ABC$ is right-angled at $C$,then the value of $\cos(A + B)$ is

Given that $\sin \theta + 2 \cos \theta = 1$,prove that $2 \sin \theta - \cos \theta = 2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module