The correct option is C √(3) cosec 20^∘ sec 20^∘=√(3)/sin 20^∘ 1/cos 20^∘ =√(3) cos 20^∘ sin 20^∘/sin 20^∘ cos 20^∘=2[√(3)/2cos 20^∘ 1/2sin 20^∘]/2/2sin 20^∘ c ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

√3cosec 20∘- ...

Question

$\sqrt{3} c o s e c 20^{\circ} - s e c 20^{\circ} =$

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Solution

The correct option is C
$\sqrt{3} c o s e c 20^{\circ} - s e c 20^{\circ} = \frac{\sqrt{3}}{s i n 20^{\circ}} - \frac{1}{c o s 20^{\circ}}$
$= \frac{\sqrt{3} c o s 20^{\circ} - s i n 20^{\circ}}{s i n 20^{\circ} c o s 20^{\circ}} = \frac{2 [\frac{\sqrt{3}}{2} c o s 20^{\circ} - \frac{1}{2} s i n 20^{\circ}]}{\frac{2}{2} s i n 20^{\circ} c o s 20^{\circ}}$
$= \frac{4 c o s (20^{\circ} + 30^{\circ})}{s i n 30^{\circ}} = \frac{4 c o s 50^{\circ}}{s i n 40^{\circ}} = \frac{4 s i n 40^{\circ}}{s i n 40^{\circ}} = 4.$

Suggest Corrections

√3 cosec 20∘−sec 20∘=

The correct option is C √3 cosec 20∘−sec 20∘=√3sin 20∘−1cos 20∘ =√3 cos 20∘−sin 20∘sin 20∘ cos 20∘=2[√32cos 20∘−12sin 20∘]22sin 20∘ cos 20∘ =4 cos(20∘+30∘)sin 30∘=4 cos 50∘sin 40∘=4 sin 40∘sin 40∘=4.

$\sqrt{3} c o s e c 20^{\circ} - s e c 20^{\circ} =$