If tan 3^{circ} + 2tan 6^{circ} + 4tan 12^{ci | vedclass

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(B) We use the identity $\cot \alpha - 	an \alpha = 2\cot 2\alpha$,which implies $	an \alpha = \cot \alpha - 2\cot 2\alpha$.Given expression: $E = \

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$\cot (15	heta)^{\circ} = 1$

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(B) We use the identity $\cot \alpha - 	an \alpha = 2\cot 2\alpha$,which implies $	an \alpha = \cot \alpha - 2\cot 2\alpha$.Given expression: $E = 	an 3^{\circ} + 2	an 6^{\circ} + 4	an 12^{\circ} + 8\cot 24^{\circ}$.Using $	an 3^{\circ} = \cot 3^{\circ} - 2\cot 6^{\circ}$:$E = (\cot 3^{\circ} - 2\cot 6^{\circ}) + 2	an 6^{\circ} + 4	an 12^{\circ} + 8\cot 24^{\circ}$.Using $2	an 6^{\circ} = 2(\cot 6^{\circ} - 2\cot 12^{\circ}) = 2\cot 6^{\circ} - 4\cot 12^{\circ}$:$E = \cot 3^{\circ} - 2\cot 6^{\circ} + 2\cot 6^{\circ} - 4\cot 12^{\circ} + 4	an 12^{\circ} + 8\cot 24^{\circ}$.Simplifying:$E = \cot 3^{\circ} - 4\cot 12^{\circ} + 4	an 12^{\circ} + 8\cot 24^{\circ}$.Using $4	an 12^{\circ} = 4(\cot 12^{\circ} - 2\cot 24^{\circ}) = 4\cot 12^{\circ} - 8\cot 24^{\circ}$:$E = \cot 3^{\circ} - 4\cot 12^{\circ} + 4\cot 12^{\circ} - 8\cot 24^{\circ} + 8\cot 24^{\circ} = \cot 3^{\circ}$.Thus,$\cot 	heta^{\circ} = \cot 3^{\circ}$,so $	heta = 3$.Checking options: $\cot (15 	imes 3)^{\circ} = \cot 45^{\circ} = 1$.

If $\tan 3^{\circ} + 2\tan 6^{\circ} + 4\tan 12^{\circ} + 8\cot 24^{\circ} = \cot \theta^{\circ}$,then:

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