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

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Question

(A) We have the expression: $	an 81^{\circ} + 	an 9^{\circ} - 	an 63^{\circ} - 	an 27^{\circ}$.Using the identity $	an(90^{\circ}-	heta) = \cot

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) We have the expression: $	an 81^{\circ} + 	an 9^{\circ} - 	an 63^{\circ} - 	an 27^{\circ}$.Using the identity $	an(90^{\circ}-	heta) = \cot 	heta$,we can write:$= (\cot 9^{\circ} + 	an 9^{\circ}) - (\cot 27^{\circ} + 	an 27^{\circ})$.Using $\cot 	heta + 	an 	heta = \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta + \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$,we get:$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.$= 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} ight)$.Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} ight) = 4 \frac{\cos 36^{\circ}}{\sin 54^{\circ}}$.Since $\sin 54^{\circ} = \cos(90^{\circ}-54^{\circ}) = \cos 36^{\circ}$,the expression simplifies to $4(1) = 4$.

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

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