√ 3 c o s e c 20 ∘ − s e c 20 ∘ = Options: A . &nbsp 2 B . &nbsp 2 s i n 20 ∘ s i n 40 ∘ C . &nbsp 4 D . &nbsp 4 s i n 20 ∘ s i n 40 ∘

Answer: Option C : C √ 3 c o s e c 20 ∘ − s e c 20 ∘ = √ 3 s i n 20 ∘ − 1 c o s 20 ∘ = √ 3 c o s 20 ∘ − s i n 20 ∘ s i n 20 ∘ c o s 20 ∘ = 2 [ √ 3 2 c o s 20 ∘ − 1 2 s i n 20 ∘ ] 2 2 s i n 20 ∘ c o s 20 ∘ = 4 c o s ( 20 ∘ + 30 ∘ ) s i n 40 ∘ = 4 c o s 50 ∘ s i n 40 ∘ = 4 s i n 40 ∘ s i n 40 ∘ = 4 .

√3cosec 20∘−sec 20∘=

Discussion Forum : Trigonometric Functions

Question -

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

Options:

A . 2

B .

\frac{2 s i n 20^{\circ}}{s i n 40^{\circ}}

C . 4

D .

\frac{4 s i n 20^{\circ}}{s i n 40^{\circ}}

Answer: Option C
:
C

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

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