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 the identities $	an(90^\circ - 	heta) = \cot 	heta$,we ha

Vedclass · Accepted Answer

$4$

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(C) Given expression: $	an 9^\circ - 	an 27^\circ - 	an 63^\circ + 	an 81^\circ$Using the identities $	an(90^\circ - 	heta) = \cot 	heta$,we have $	an 63^\circ = \cot 27^\circ$ and $	an 81^\circ = \cot 9^\circ$.Substituting these into the expression:$= (	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}$$= 2 \left( \frac{\sin 54^\circ - \sin 18^\circ}{\sin 18^\circ \sin 54^\circ} ight)$Using the sum-to-product formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$= 2 \left( \frac{2 \cos 36^\circ \sin 18^\circ}{\sin 18^\circ \sin 54^\circ} ight)$Since $\sin 54^\circ = \cos 36^\circ$,the expression simplifies to:$= 2 	imes 2 = 4$

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

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