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(A

Vedclass · Accepted Answer

$4$

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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}$:$	an 81^{\circ} + 	an 9^{\circ} = \frac{\sin(81^{\circ}+9^{\circ})}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{\sin 90^{\circ}}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} = \frac{2}{\sin 18^{\circ}}$Similarly,$	an 63^{\circ} + 	an 27^{\circ} = \frac{\sin(63^{\circ}+27^{\circ})}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{\sin 90^{\circ}}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} = \frac{2}{\sin 54^{\circ}}$Now,the expression becomes $\frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}} = 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}} ight)$Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$\sin 54^{\circ} - \sin 18^{\circ} = 2 \cos 36^{\circ} \sin 18^{\circ}$So,$2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}} ight) = 2 	imes 2 = 4$

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

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