A) 1
B) 2
C) 3
D) \[\sqrt{3}/2\]
Correct Answer: C
Solution :
\[\tan \,\,{{20}^{o}}\tan \,\,{{40}^{o}}\tan \,\,{{60}^{o}}\tan \,\,{{80}^{o}}\] \[=\frac{\sin \,\,{{20}^{o}}\sin \,\,{{40}^{o}}\sin \,\,{{80}^{o}}\tan {{60}^{o}}}{\cos \,\,{{20}^{o}}\cos \,\,{{40}^{o}}\cos \,\,{{80}^{o}}}\] Here \[{{N}^{r}}=(\sin \,\,{{20}^{o}}\sin \,\,{{40}^{o}}\sin \,\,{{80}^{o}})\] \[=\frac{\sin \,\,{{20}^{o}}}{2}\,(2\,\,\sin \,\,{{40}^{o}}\sin \,\,{{80}^{o}})\] \[=\frac{\sin \,\,{{20}^{o}}}{2}\,(\cos \,\,{{40}^{o}}-\cos \,\,{{120}^{o}})\] \[=\frac{1}{2}\sin \,\,{{20}^{o}}\,\left( 1-2\,\,{{\sin }^{2}}{{20}^{o}}+\frac{1}{2} \right)\] \[=\frac{1}{2}\sin \,\,{{20}^{o}}\,\left( \frac{3}{2}-2\,\,{{\sin }^{2}}{{20}^{o}} \right)=\frac{\sin \,{{60}^{o}}}{4}=\frac{\sqrt{3}}{8}\] Now, we take \[{{D}^{r}}=\cos {{20}^{o}}\cos {{40}^{o}}\cos {{80}^{o}}\] \[=\frac{\sin \,\,{{2}^{3}}\,{{20}^{o}}}{{{2}^{3}}\,\sin \,\,{{20}^{o}}}=\frac{\sin \,\,{{160}^{o}}}{8\,\,\sin \,\,{{20}^{o}}}=\frac{\sin \,\,{{20}^{o}}}{8\,\,\sin \,\,{{20}^{o}}}=\frac{1}{8}\] \[\therefore \] Hence \[\tan \,\,{{20}^{o}}\tan \,\,{{40}^{o}}\tan \,\,{{80}^{o}}=\frac{\sqrt{3}/8}{1/8}\] Therefore\[\tan {{20}^{o}}\tan {{40}^{o}}\tan {{60}^{o}}\tan {{80}^{o}}=\sqrt{3}.\sqrt{3}=3\].
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