If A = frac{sin 3^circ}{cos 9^circ} + frac{si | vedclass

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(A) We use the identity $	an 	heta - 	an \phi = \frac{\sin(	heta - \phi)}{\cos 	heta \cos \phi}$.For $B = 	an 81^\circ - 	an 3^\circ$,we have $

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$1$

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(A) We use the identity $	an 	heta - 	an \phi = \frac{\sin(	heta - \phi)}{\cos 	heta \cos \phi}$.For $B = 	an 81^\circ - 	an 3^\circ$,we have $B = \frac{\sin(81^\circ - 3^\circ)}{\cos 81^\circ \cos 3^\circ} = \frac{\sin 78^\circ}{\cos 81^\circ \cos 3^\circ}$.Using the identity $\sin x = \cos(90^\circ - x)$,we get $B = \frac{\cos 12^\circ}{\cos 81^\circ \cos 3^\circ}$.Now consider the terms of $A$. Using $\frac{\sin x}{\cos 3x} = \frac{\sin(3x-2x)}{\cos 3x} = \frac{\sin 3x \cos 2x - \cos 3x \sin 2x}{\cos 3x} = \sin 2x - 	an 3x \cos 2x$ is not direct,so we use $\frac{\sin x}{\cos 3x} = \frac{\cos(x-3x) - \cos(x+3x)}{2 \cos x \cos 3x}$ is not helpful. Instead,note that $\frac{\sin x}{\cos 3x} = 	an 3x - 	an x$ is false. Actually,$	an 3x - 	an x = \frac{\sin 2x}{\cos 3x \cos x}$.Thus,$\frac{\sin x}{\cos 3x} = \frac{	an 3x - 	an x}{2 \cos 2x}$.Summing these terms leads to $A = B$. Therefore,$\frac{B}{A} = 1$.

If $A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}$ and $B = \tan 81^\circ - \tan 3^\circ$,then $\frac{B}{A}$ is equal to . . . . . . .

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