If tan {3^o} + 2tan {6^o} + 4tan {12^o} + 8co | vedclass

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

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$\cot (15	heta)^o = 1$

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

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

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