tan 9^circ - tan 27^circ - tan 63^circ + tan | vedclass

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Question

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

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$4$

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(C) Given expression: $	an 9^\circ - 	an 27^\circ - 	an 63^\circ + 	an 81^\circ$Using $	an(90^\circ - 	heta) = \cot 	heta$,we have $	an 63^\circ = \cot 27^\circ$ and $	an 81^\circ = \cot 9^\circ$.Expression $= (	an 9^\circ + \cot 9^\circ) - (	an 27^\circ + \cot 27^\circ)$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}$.Expression $= \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ} = 2 \left( \frac{\sin 54^\circ - \sin 18^\circ}{\sin 18^\circ \sin 54^\circ} ight)$Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get $\sin 54^\circ - \sin 18^\circ = 2 \cos 36^\circ \sin 18^\circ$.Expression $= 2 \left( \frac{2 \cos 36^\circ \sin 18^\circ}{\sin 18^\circ \sin 54^\circ} ight) = 4 \frac{\cos 36^\circ}{\sin 54^\circ}$.Since $\sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ$,the expression $= 4(1) = 4$.

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

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