A) \[1\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\sqrt{3}\]
D) \[\frac{\sqrt{3}-1}{\sqrt{3}+1}\]
E) \[\frac{\sqrt{3}+1}{\sqrt{3}-1}\]
Correct Answer: C
Solution :
Let \[\tan 4A=\sqrt{3}=\tan \frac{\pi }{3}\] \[\Rightarrow \] \[A=\frac{\pi }{12}\] \[\therefore \] \[sin\text{ }4A-cos\text{ }2A=sin\frac{\pi }{3}-cos\frac{\pi }{6}\] \[=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=0\] and \[\cos 4A-\sin 2A=\sin \frac{\pi }{3}-\sin \frac{\pi }{6}\] \[=\frac{1}{2}-\frac{1}{2}=0\] \[\therefore \] \[sin\text{ }4A-cos\text{ }2A=cos\text{ }4A-sin\text{ }2A\] Hence, our assumption is true. \[\therefore \] \[\tan 4A=\sqrt{3}\]
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