A) 2
B) 3
C) 4
D) None of these
Correct Answer: B
Solution :
We have, \[{{\tan }^{6}}40{}^\circ -33{{\tan }^{4}}40{}^\circ +27{{\tan }^{2}}40{}^\circ \]| \[\Rightarrow \]\[\tan 3(40{}^\circ )=\frac{3\tan 40{}^\circ -{{\tan }^{3}}40{}^\circ }{1-3{{\tan }^{2}}40{}^\circ }\] |
| \[\Rightarrow \]\[\tan 120{}^\circ =\frac{3\tan 40{}^\circ -{{\tan }^{3}}40{}^\circ }{1-3{{\tan }^{2}}40{}^\circ }\] |
| \[\Rightarrow \]\[-\sqrt{3}(1-3{{\tan }^{2}}40{}^\circ )=3\tan 40{}^\circ -{{\tan }^{3}}40{}^\circ \] |
| \[\Rightarrow \]\[3\,{{(1-3{{\tan }^{2}}40{}^\circ )}^{2}}\] |
| \[={{\tan }^{2}}40{}^\circ {{(3-{{\tan }^{2}}40{}^\circ )}^{2}}\] |
| \[\Rightarrow \]\[{{\tan }^{6}}40{}^\circ -33{{\tan }^{4}}40{}^\circ +27{{\tan }^{2}}60{}^\circ =3\] |
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