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Question

Mathematics Question on Trigonometric Identities

1cos⁡80∘−3sin⁡80∘\frac{1}{\cos 80{}^\circ }-\frac{\sqrt{3}}{\sin 80{}^\circ } is equal to:

A

2\sqrt{2}

B

3\sqrt{3}

C

2

D

4

Answer

4

Explanation

Solution

sin⁡80∘−cos⁡80∘3sin⁡80∘cos⁡80∘\frac{\sin 80{}^\circ -\cos 80{}^\circ \sqrt{3}}{\sin 80{}^\circ \cos 80{}^\circ }
=2[12sin⁡80∘−32cos⁡80∘]sin⁡80∘cos⁡80∘=\frac{2\left[ \frac{1}{2}\sin 80{}^\circ -\frac{\sqrt{3}}{2}\cos 80{}^\circ \right]}{\sin 80{}^\circ \cos 80{}^\circ }
=4[sin⁡(80∘−60∘)]2sin⁡80∘cos⁡80∘=\frac{4[\sin (80{}^\circ -60{}^\circ )]}{2\sin 80{}^\circ \cos 80{}^\circ }
=4sin⁡20∘2sin⁡80∘=4sin⁡20∘sin⁡(π−20∘)=4=\frac{4\sin 20{}^\circ }{2\sin 80{}^\circ }=\frac{4\sin 20{}^\circ }{\sin (\pi -20{}^\circ )}=4