A) \[tan\text{ }50{}^\circ \]
B) \[tan\text{ 6}0{}^\circ \]
C) \[tan\text{ 3}0{}^\circ \]
D) None of these
Correct Answer: B
Solution :
\[\tan \,20{}^\circ \,\tan 40{}^\circ \,\tan 80{}^\circ \] \[=\tan 20{}^\circ \,\tan \left( 60{}^\circ -20{}^\circ \right)\tan \left( 60{}^\circ +20{}^\circ \right)\] \[=\tan 20{}^\circ \times \frac{\tan 60{}^\circ -\tan 20{}^\circ }{1+\tan 60{}^\circ \times \tan 20{}^\circ }\] \[\times \frac{\tan \,60{}^\circ +\tan 20{}^\circ }{1-\tan 60{}^\circ \,\tan 20{}^\circ }\] \[=\tan 20{}^\circ \times \frac{\sqrt{3}-\tan 20{}^\circ }{1+\sqrt{3}\,\tan 20{}^\circ }\times \frac{\sqrt{3}+\tan 20{}^\circ }{1-\sqrt{3}\,\tan 20{}^\circ }\] \[=\tan 20{}^\circ \times \frac{3-{{\tan }^{2}}20{}^\circ }{1-3{{\tan }^{2}}20{}^\circ }\] \[=\frac{3\tan 20{}^\circ -{{\tan }^{3}}20{}^\circ }{1-3{{\tan }^{2}}20{}^\circ }\] \[\tan 3\times 20{}^\circ =\]
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