A) \[a<\frac{1}{2}\]
B) \[a>\frac{1}{2}\]
C) \[a<0\]
D) \[a>0\]
Correct Answer: A
Solution :
\[\sqrt{3}\cot 20{}^\circ -4\cos 20{}^\circ \] \[=\frac{\sqrt{3}\cos 20{}^\circ }{\sin 20{}^\circ }-4\cos 20{}^\circ \] \[=\frac{\sqrt{3}\cos 20{}^\circ -4\sin 20{}^\circ \cos 20{}^\circ }{\sin 20{}^\circ }\] \[=\frac{2.\sin 60{}^\circ \cos 20{}^\circ -2\sin 40{}^\circ }{\sin 20{}^\circ }\] \[[\because \sqrt{3}=2.\frac{\sqrt{3}}{2}=2\sin 60{}^\circ \]and \[2\sin 20{}^\circ \cos 20{}^\circ =\sin 40{}^\circ ]\] \[=\frac{\sin 80{}^\circ +\sin 40{}^\circ -2\sin 40{}^\circ }{\sin 20{}^\circ }\] [using\[2\text{ }sin\text{ }A\text{ }cos\text{ }B\] \[=sin(A+B)+sin(A-B)]\] \[=\frac{\sin 80{}^\circ -\sin 40{}^\circ }{\sin 20{}^\circ }\] \[=\frac{2\cos 60{}^\circ \sin 20{}^\circ }{\sin 20{}^\circ }=2.\frac{1}{2}=1\] [using \[sin\text{ }C-sin\text{ }D=2\cos \frac{C+D}{2}.\sin \frac{C-D}{2}\]]
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