Question

$\sqrt{3}\, cosec \,20^{\circ} - \sec\, 20^{\circ}$ =

• A. 4
• B. 2
• C. 1
• D. 3

Answer 1

The Correct Option is A

$\sqrt{3} cos e 20^{\circ}-sec 20^{\circ}$
$=\frac{\sqrt{3}}{sin \,20^{\circ} }-\frac{1}{cos\, 20^{\circ}}$
$=\frac{\sqrt{3}cos 20^{\circ}-sin\,20^{\circ}}{sin\,20^{\circ}. cos\,20^{\circ}}$
$=\frac{2\left[\frac{\sqrt{3}}{2} cos\,20^{\circ}-\frac{1}{2} sin\, 20^{\circ}\right]}{\frac{1}{2}\left[2 sin\, 20^{\circ}. cos\, 20^{\circ}\right]}$
$= 4\frac{\left[sin\,60^{\circ}. cos\,20^{\circ}-cos\,60^{\circ}. sin\,20^{\circ}\right]}{sin \, 40^{\circ}}$
$=4\frac{sin\left(60^{\circ}-20^{\circ}\right)}{sin\, 40^{\circ}}$
$=4\frac{sin\,40^{\circ}}{sin\,40^{\circ}}$
$=4$