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

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Question

(D) हमारे पास है,$	an 81^{\circ} - 	an 63^{\circ} - 	an 27^{\circ} + 	an 9^{\circ}$$= (	an 81^{\circ} + 	an 9^{\circ}) - (	an 63^{\circ} + 	an

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) हमारे पास है,$	an 81^{\circ} - 	an 63^{\circ} - 	an 27^{\circ} + 	an 9^{\circ}$$= (	an 81^{\circ} + 	an 9^{\circ}) - (	an 63^{\circ} + 	an 27^{\circ})$$= (\cot 9^{\circ} + 	an 9^{\circ}) - (\cot 27^{\circ} + 	an 27^{\circ})$$= \left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}} + \frac{\sin 9^{\circ}}{\cos 9^{\circ}}ight) - \left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}} + \frac{\sin 27^{\circ}}{\cos 27^{\circ}}ight)$$= \left(\frac{\cos^2 9^{\circ} + \sin^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}ight) - \left(\frac{\cos^2 27^{\circ} + \sin^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}ight)$$= \left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}ight) - \left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}ight)$$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$$= 2 \left(\frac{1}{\sin 18^{\circ}} - \frac{1}{\cos 36^{\circ}}ight)$$= 2 \left(\frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1}ight)$$= 8 \left(\frac{\sqrt{5} + 1 - (\sqrt{5} - 1)}{5 - 1}ight)$$= 8 \left(\frac{2}{4}ight) = 4$

$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ का मान है

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