The expression tan 9^{circ}-tan 27^{circ}-tan | vedclass

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(A) We know that $	an 63^{\circ} = \cot 27^{\circ}$ and $	an 81^{\circ} = \cot 9^{\circ}$.Substituting these into the expression:$	an 9^{\circ} - \

Vedclass · Accepted Answer

$4$

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(A) We know that $	an 63^{\circ} = \cot 27^{\circ}$ and $	an 81^{\circ} = \cot 9^{\circ}$.Substituting these into the expression:$	an 9^{\circ} - 	an 27^{\circ} - \cot 27^{\circ} + \cot 9^{\circ} = (	an 9^{\circ} + \cot 9^{\circ}) - (	an 27^{\circ} + \cot 27^{\circ})$Using the identity $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$:$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$Given $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$:$= \frac{2}{(\sqrt{5}-1)/4} - \frac{2}{(\sqrt{5}+1)/4} = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1}$$= 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} ight) = 8 \left( \frac{2}{5-1} ight) = 8 \left( \frac{2}{4} ight) = 4$.

The expression $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$ is equal to

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