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  4. The value of √3 cot 20° -4 cos 20° is equal to

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

JamiaJamia 2010

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} $ ]