$$\tan81^\circ-\tan63^\circ-\tan27^\circ+\tan9^\circ$$ At first, I tried to break the tangent into multiple angles… $$\text{FAILED}$$ Then I proceeded by converting it into other trigonometric ratios… $$\text{FAILED}$$ Please suggest me an approach to such questions. A solution is most welcomed.
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Rearrange the above expression - $\tan 81 + \tan 9 - (\tan 63 + \tan 27)$. which formula should u use now? – Dark_Knight Oct 14 '16 at 07:15
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HINT: We have $$\tan(u) + \tan(v) = {\sin(u) \over \cos(u)} + {\sin(v) \over \cos(v)} = {\sin(u)\cos(v) + \cos(u)\sin(v) \over \cos(u)\cos(v)} = {\sin(u + v) \over \cos(u)\cos(v)}$$
Can you see what to do with $\tan(u) - \tan(v)$?
Decaf-Math
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\begin{align*} \tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ} &= \tan (90^{\circ}-9^{\circ})-\tan (90^{\circ}-27^{\circ})- \tan 27^{\circ}+\tan 9^{\circ} \\ &= \cot 9^{\circ}-\cot 27^{\circ}-\tan 27^{\circ}+\tan 9^{\circ} \\ &= \frac{\cos 9^{\circ}}{\sin 9^{\circ}}+ \frac{\sin 9^{\circ}}{\cos 9^{\circ}}- \frac{\cos 27^{\circ}}{\sin 27^{\circ}}- \frac{\sin 27^{\circ}}{\cos 27^{\circ}} \\ &= \frac{\cos^2 9^{\circ}+\sin^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}- \frac{\cos^2 27^{\circ}+\sin^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}} \\ &= 2(1+\sqrt{5})-2(\sqrt{5}-1) \\ &= 4 \end{align*}
Ng Chung Tak
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