Thursday, March 26, 2015

2015/023) show that $\tan\, 9^{\circ} - \tan\, 27^{\circ}- \tan\, 63^{\circ} + \tan\, 81^{\circ}= 4$


we have $\tan\, x + \cot\, x = \tan\, x + \dfrac{1}{\tan\, x}$
 = $\dfrac{tan ^2 x + 1}{\ tan\, x} = \dfrac{sec^2 x}{\tan,x}= \dfrac{1}{\cos\,x\sin\,x}=\dfrac{2}{2\cos\,x\sin\,x} = \dfrac{2}{\sin 2x} = 2 \csc 2x$

 so
 $\tan\, 9^{\circ} + \tan\, 81^{\circ} = 2 \csc\, 18^{\circ} $
$\tan\, 27^{\circ} + \tan\, 63^{\circ} = 2 \csc\,  54^{\circ} $

so given value

= $\tan\, 9^{\circ} - \tan\, 27^{\circ}- \tan\, 63^{\circ} + \tan\, 81^{\circ}=2( \csc\, 18^{\circ} - \csc\, 54^{\circ})$ 

= $2(\dfrac{1}{\sin\,18^{\circ}} - \dfrac{1}{\sin\, 54^{\circ}})$

= $2 \dfrac{2 \cos\, 36^{\circ}  \sin\, 18^{\circ}}{ \sin \,18^{\circ}  \sin \,54^{\circ} }$
4

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