Question

The value of $\sqrt{3}\cot 20{}^{\circ }-4\cos 20{}^{\circ }$ is equal to

• A. 1
• B. $-1$
• C. 0
• D. None of these

Answer 1

Answer: (A)

$\sqrt{3}\cot 20{}^{\circ }-4\cos 20{}^{\circ }$ $=\frac{\sqrt{3}\cos 20{}^{\circ }}{\sin 20{}^{\circ }}-4\cos 20{}^{\circ }$ $=\frac{\sqrt{3}\cos 20{}^{\circ }-4\sin 20{}^{\circ }\cos 20{}^{\circ }}{\sin 20{}^{\circ }}$ $=\frac{2.\sin 60{}^{\circ }\cos 20{}^{\circ }-2\sin 40{}^{\circ }}{\sin 20{}^{\circ }}$ $[\because \sqrt{3}=2.\frac{\sqrt{3}}{2}=2\sin 60{}^{\circ }$ 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 $2sinAcosB$ $=sin(A+B)+sin(A-B)]$ $=\frac{\sin 80{}^{\circ }-\sin 40{}^{\circ }}{\sin 20{}^{\circ }}$ $=\frac{2\cos 60{}^{\circ }\sin 20{}^{\circ }}{\sin 20{}^{\circ }}=2.\frac{1}{2}=1$ [using $sinC-sinD=2\cos \frac{C+D}{2}.\sin \frac{C-D}{2}$ ]