-3cosec 20- 20-IIT 1988-

The correct option is C 4 √(3) cosec 20^∘ sec 20^∘=√(3)/sin 20^∘ 1/cos 20^∘ =√(3) cos20^∘ sin20^∘/sin20^∘ cos20^∘=2[√(3) 2cos20^∘ 1 2sin20^∘]/22sin20^∘ cos20^∘ ...

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Standard XI

Mathematics

Trigonometric Ratios of Compound Angles

√3cosec 20∘- ...

Question

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

[IIT 1988]

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Solution

The correct option is C
4

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

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

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

is equal to :

Q. The value of

\sqrt{3} cosec 20^{\circ} - sec 20^{\circ}

Q. The value of

\sqrt{3} cosec 20^{\circ} - sec 20^{\circ}

is equal to

Q. The value of the expression

\sqrt{3} cosec 20^{\circ} - sec 20^{\circ}

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√3cosec 20∘−sec 20∘= [IIT 1988]

The correct option is C 4 √3cosec 20∘−sec 20∘=√3sin 20∘−1cos 20∘ =√3 cos20∘−sin20∘sin20∘ cos20∘=2[√32cos20∘−12sin20∘]22sin20∘ cos20∘ =4cos(20∘+30∘)sin40∘=4cos50∘sin40∘=4sin40∘sin40∘=4.

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

[IIT 1988]