The value of 1sin20-1-3cos20- is

Given: nbsp;1sin20^∘ 1√(3)cos20^∘ =√(3)cos20^∘ sin20^∘√(3)cos20^∘sin20^∘ = √(3)2cos20^∘ 12sin20^∘√(3)4·(2cos20^∘sin20^∘) = 4√(3)·sin60^∘cos20^∘ cos60^∘sin20^∘ ...

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Standard XII

Mathematics

Definition of Limit

The value of ...

Question

The value of $\frac{1}{sin 20^{\circ}} - \frac{1}{\sqrt{3} cos 20^{\circ}}$ is

4

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4 \sqrt{3}

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\frac{1}{4}

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\frac{4}{\sqrt{3}}

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Solution

The correct option is D $\frac{4}{\sqrt{3}}$
Given: $\frac{1}{sin 20^{\circ}} - \frac{1}{\sqrt{3} cos 20^{\circ}}$
$= \frac{\sqrt{3} cos 20^{\circ} - sin 20^{\circ}}{\sqrt{3} cos 20^{\circ} sin 20^{\circ}}$
$= \frac{\frac{\sqrt{3}}{2} cos 20^{\circ} - \frac{1}{2} sin 20^{\circ}}{\frac{\sqrt{3}}{4} \cdot (2 cos 20^{\circ} sin 20^{\circ})}$
$= \frac{4}{\sqrt{3}} \cdot \frac{sin 60^{\circ} cos 20^{\circ} - cos 60^{\circ} sin 20^{\circ}}{sin 40^{\circ}}$
$= \frac{4}{\sqrt{3}} \cdot \frac{sin 40^{\circ}}{sin 40^{\circ}}$
$= \frac{4}{\sqrt{3}}$

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The value of 1sin20∘−1√3cos20∘ is

The correct option is D 4√3Given: 1sin20∘−1√3cos20∘ =√3cos20∘−sin20∘√3cos20∘sin20∘ =√32cos20∘−12sin20∘√34⋅(2cos20∘sin20∘) =4√3⋅sin60∘cos20∘−cos60∘sin20∘sin40∘ =4√3⋅sin40∘sin40∘ =4√3

The value of $\frac{1}{sin 20^{\circ}} - \frac{1}{\sqrt{3} cos 20^{\circ}}$ is