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

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Question

(c) $	an \,\,{9^o} - 	an \,\,{27^o} - 	an \,\,{63^o} + 	an \,\,{81^o}$

$ = 	an \,\,{9^o} - 	an \,\,{27^o} - \cot \,\,{27^o} + \cot \,\,{9^o}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c) $	an \,\,{9^o} - 	an \,\,{27^o} - 	an \,\,{63^o} + 	an \,\,{81^o}$

$ = 	an \,\,{9^o} - 	an \,\,{27^o} - \cot \,\,{27^o} + \cot \,\,{9^o}$

$ = (	an \,\,{9^o} + \cot \,\,{9^o}) - (	an \,\,{27^o} + \cot \,\,{27^o})$

$ = \frac{{\cos ({9^o} - {9^o})}}{{\sin {9^o}\cos {9^o}}} - \frac{{\cos ({{27}^o} - {{27}^o})}}{{\sin {{27}^o}.\cos {{27}^o}}} = \frac{2}{{\sin {{18}^o}}} - \frac{2}{{\sin {{54}^o}}}$

$ = 2\,\left\{ {\frac{{\sin \,\,{{54}^o} - \sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}\,\sin \,\,{{54}^o}}}} ight\} $

$= 2.\,\,\frac{{2\,.\,\cos \,\,{{36}^o}.\,\sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}.\,\sin \,\,{{54}^o}}} = 4$

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

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