The expression tan 9^{circ}-tan 27^{circ}-tan | vedclass

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Question

(A) Given expression: $	an 9^{\circ}-	an 27^{\circ}-	an 63^{\circ}+	an 81^{\circ}$ Using $	an 63^{\circ} = \cot 27^{\circ}$ and $	an 81^{\circ}

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given expression: $	an 9^{\circ}-	an 27^{\circ}-	an 63^{\circ}+	an 81^{\circ}$ Using $	an 63^{\circ} = \cot 27^{\circ}$ and $	an 81^{\circ} = \cot 9^{\circ}$,we get: $(	an 9^{\circ} + \cot 9^{\circ}) - (	an 27^{\circ} + \cot 27^{\circ})$ $= \left(\frac{\sin 9^{\circ}}{\cos 9^{\circ}} + \frac{\cos 9^{\circ}}{\sin 9^{\circ}}ight) - \left(\frac{\sin 27^{\circ}}{\cos 27^{\circ}} + \frac{\cos 27^{\circ}}{\sin 27^{\circ}}ight)$ $= \frac{\sin^2 9^{\circ} + \cos^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}} - \frac{\sin^2 27^{\circ} + \cos^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}$ $= \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} - \frac{1}{\sin 27^{\circ} \cos 27^{\circ}}$ $= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$ Substituting $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$: $= \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = \frac{8(\sqrt{5}+1) - 8(\sqrt{5}-1)}{5-1} = \frac{16}{4} = 4$

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

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