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Let $$A=\dfrac{1}{\sin20^\circ\cos40^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ\cos40^\circ}$$ and $$B=\cot40^\circ+\cot230^\circ-\tan185^\circ(\tan230^\circ-\cot50^\circ)$$ Find $AB$.

My attempt : \begin{align*} A&=\dfrac{1}{\cos40^\circ}(\dfrac{1}{\sin20^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ})\\ &=\dfrac{2(\sqrt{3}\cos20^\circ+\sin20^\circ)}{\sqrt{3}\cos^2 40^\circ}\\ B&=\tan50^\circ(1-\tan5^\circ)+\tan40^\circ(1+\tan5^\circ) \end{align*} I tried to factor $A$ and $B$ because I think it will cancel each other out.

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    $\begingroup$ Your $A$ is wrong. $\left(\dfrac{1}{\sin20^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ}\right) =\dfrac{2(\sqrt{3}\cos20^\circ+\sin20^\circ)}{\sqrt{3}\color{blue}\sin 40^\circ}$ $\endgroup$ – Alexey Burdin Aug 8 '20 at 3:47
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\begin{align*} A &= \frac{4(\sin 60^{\circ} \cos 20^{\circ} + \cos 60^{\circ} \sin 20^{\circ})}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}} = \frac{4\sin 80^{\circ}}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}} \\&= \frac{8\sin 80^{\circ}}{\sqrt{3}\sin 80^{\circ}} = \frac{8\sqrt{3}}{3}\\ B &= \cot 40^{\circ} + \cot 230^{\circ} - \tan 185^{\circ}(\tan 230^{\circ} - \cot 50^{\circ}) \\ &=\cot 40^{\circ} + \tan 40^{\circ} - \tan 5^{\circ}(\tan 50^{\circ} - \cot 50^{\circ}) \\ &= \cot 40^{\circ} + \tan 40^{\circ} - \tan 5^{\circ}(\cot 40^{\circ} - \tan 40^{\circ})\\ \therefore AB \tan 40^{\circ} &=\frac{8\sqrt{3}}{3} \left ( 1 +\tan^{2} 40^{\circ} - \tan 5^{\circ} ( 1 - \tan^{2} 40^{\circ}) \right ) \\&= \frac{8\sqrt{3}}{3} \left ( 1 +\tan^{2} 40^{\circ} - (\frac{1- \tan 40^{\circ}}{1 + \tan 40^{\circ}}) ( 1 - \tan^{2} 40^{\circ}) \right ) \\ &= \frac{8\sqrt{3}}{3} \left ( 1 +\tan^{2} 40^{\circ} - ( 1 - \tan 40^{\circ})^{2} \right ) \\ &= \frac{16\sqrt{3}}{3} \tan 40^{\circ} \end{align*}

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  • $\begingroup$ Isn't $A= \frac{2(\sin 60^{\circ} \cos 20^{\circ} + \cos 60^{\circ} \sin 20^{\circ})}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}}$ ? $\endgroup$ – Ken Aug 8 '20 at 4:04

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