The value of $\tan 40^{0} + \tan 11^{0} + \tan 20^{0} - \tan 56^{0} + \tan 56^{0} \tan 11^{0} + \sqrt{3} \tan 40^{0} \tan 20^{0} =$

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$\sqrt{3} - 1$

$\sqrt{3} + 1$

answer is A.

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Detailed Solution

$\tan 60^{\circ} = \tan (40^{\circ} + 20^{\circ}) \sqrt{3} = \frac{\tan 40^{\circ} + \tan 20^{\circ}}{1 - \tan 40^{\circ} \tan 20^{\circ}} \sqrt{3} - \sqrt{3} \tan 40^{\circ} \tan 20^{\circ} = \tan 40^{\circ} + \tan 2^{\circ} \tan 40^{\circ} + \tan 20^{\circ} + \sqrt{3} \tan 40^{\circ} \tan 20^{\circ} = \sqrt{3}$
$\tan 45^{\circ} = \tan (56^{\circ} - 11^{\circ}) 1 = \frac{\tan 56^{o} - \tan 11^{o}}{1 + \tan 56^{o} \tan 11^{o}} \tan 56^{o} \tan 11^{o} - \tan 56^{o} + \tan 11^{o} = - 1$
$\tan 40^{0} + \tan 11^{0} + \tan 20^{0} - \tan 56^{0} + \tan 56^{0} \tan 11^{0} + \sqrt{3} \tan 40^{0} \tan 20^{0} = \sqrt{3} - 1$