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

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Question

(D) We have the expression: $	an 9^{\circ}-	an 27^{\circ}-	an 63^{\circ}+	an 81^{\circ}$ Rearranging the terms: $(	an 81^{\circ}+	an 9^{\circ})

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We have the expression: $	an 9^{\circ}-	an 27^{\circ}-	an 63^{\circ}+	an 81^{\circ}$ Rearranging the terms: $(	an 81^{\circ}+	an 9^{\circ}) - (	an 63^{\circ}+	an 27^{\circ})$ Using the identity $	an(90^{\circ}-	heta) = \cot 	heta$: $= (\cot 9^{\circ}+	an 9^{\circ}) - (\cot 27^{\circ}+	an 27^{\circ})$ Using $\cot 	heta + 	an 	heta = \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta + \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$: $= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$ $= 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} ight)$ Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$: $= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} ight)$ Since $\sin 54^{\circ} = \cos 36^{\circ}$: $= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\cos 36^{\circ} \sin 18^{\circ}} ight) = 2 	imes 2 = 4$

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

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