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\(\frac{{3{\rm{tan}}{{20}^0} - {\rm{ta}}{{\rm{n}}^3}{{20}^0}}}{{1 - 3{\rm{ta}}{{\rm{n}}^3}{{20}^0}}}\) ?

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\(\frac{{3{\rm{tan}}{{20}^0} - {\rm{ta}}{{\rm{n}}^3}{{20}^0}}}{{1 - 3{\rm{ta}}{{\rm{n}}^3}{{20}^0}}}\) ?
1). \(\frac{1}{{\sqrt 3 }}\)
2). 1
3). \(\sqrt 3 \)
4). \(\infty \)


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1 vote
Answered by on | Votes 1 |

$(\frac{{3{\rm{tan}}{{20}^0} - {\rm{ta}}{{\rm{n}}^3}{{20}^0}}}{{1 - 3{\rm{ta}}{{\rm{n}}^3}{{20}^0}}})$

[tan3A $( = \frac{{3tanA - ta{n^3}A}}{{1 - 3ta{n^3}A}})$]

tan3 × 20°

∴ tan 60° $( = \sqrt 3 )$

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