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Question

Mathematics Question on Trigonometric Functions

3 cosec 20∘−sec⁡ 20∘\sqrt{3}\, cosec \,20^{\circ} - \sec\, 20^{\circ} =

A

4

B

2

C

1

D

3

Answer

4

Explanation

Solution

3cose20∘−sec20∘\sqrt{3} cos e 20^{\circ}-sec 20^{\circ}
=3sin 20∘−1cos 20∘=\frac{\sqrt{3}}{sin \,20^{\circ} }-\frac{1}{cos\, 20^{\circ}}
=3cos20∘−sin 20∘sin 20∘.cos 20∘=\frac{\sqrt{3}cos 20^{\circ}-sin\,20^{\circ}}{sin\,20^{\circ}. cos\,20^{\circ}}
=2[32cos 20∘−12sin 20∘]12[2sin 20∘.cos 20∘]=\frac{2\left[\frac{\sqrt{3}}{2} cos\,20^{\circ}-\frac{1}{2} sin\, 20^{\circ}\right]}{\frac{1}{2}\left[2 sin\, 20^{\circ}. cos\, 20^{\circ}\right]}
=4[sin 60∘.cos 20∘−cos 60∘.sin 20∘]sin 40∘= 4\frac{\left[sin\,60^{\circ}. cos\,20^{\circ}-cos\,60^{\circ}. sin\,20^{\circ}\right]}{sin \, 40^{\circ}}
=4sin(60∘−20∘)sin 40∘=4\frac{sin\left(60^{\circ}-20^{\circ}\right)}{sin\, 40^{\circ}}
=4sin 40∘sin 40∘=4\frac{sin\,40^{\circ}}{sin\,40^{\circ}}
=4=4