tan 81^{circ}-tan 63^{circ}-tan 27^{circ}+tan | vedclass

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(D) We have the expression: $	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$.Using the identity $	an A + 	an B = \frac{\sin(

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$4$

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

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$ is equal to

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