The value of tan 9^{circ} - tan 27^{circ} - t | vedclass

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(C) Given expression: $E = 	an 9^{\circ} - 	an 27^{\circ} - 	an 63^{\circ} + 	an 81^{\circ}$Using $	an(90^{\circ} - 	heta) = \cot 	heta$,we hav

Vedclass · Accepted Answer

$4$

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(C) Given expression: $E = 	an 9^{\circ} - 	an 27^{\circ} - 	an 63^{\circ} + 	an 81^{\circ}$Using $	an(90^{\circ} - 	heta) = \cot 	heta$,we have $	an 81^{\circ} = \cot 9^{\circ}$ and $	an 63^{\circ} = \cot 27^{\circ}$.So,$E = (	an 9^{\circ} + \cot 9^{\circ}) - (	an 27^{\circ} + \cot 27^{\circ})$.Using $	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}$.$E = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.We know $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.$E = \frac{2 	imes 4}{\sqrt{5}-1} - \frac{2 	imes 4}{\sqrt{5}+1} = 8 \left( \frac{1}{\sqrt{5}-1} - \frac{1}{\sqrt{5}+1} ight)$.$E = 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} ight) = 8 \left( \frac{2}{5-1} ight) = 8 \left( \frac{2}{4} ight) = 4$.

The value of $\tan 9^{\circ} - \tan 27^{\circ} - \tan 63^{\circ} + \tan 81^{\circ}$ is $............$.

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