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Question: The value of \[\sqrt {{\mathbf{3}}} {\mathbf{cosec2}}{{\mathbf{0}}^ \circ } - {\mathbf{sec2}}{{\math...

The value of 3cosec20∘−sec20∘  \sqrt {{\mathbf{3}}} {\mathbf{cosec2}}{{\mathbf{0}}^ \circ } - {\mathbf{sec2}}{{\mathbf{0}}^ \circ }\; is
A.2
B.2sin⁡20∘sin⁡40∘\dfrac{{2\sin 20^\circ }}{{\sin 40^\circ }}
C.4
D.4sin⁡20∘sin⁡40∘\dfrac{{4\sin 20^\circ }}{{\sin 40^\circ }}

Explanation

Solution

Hint : To find the value of this trigonometric expression we will first convert the given trigonometric ratios into sin and cos. Then take LCM and solve. We will get the result in the form of sin(A−B)sin\left( {A - B} \right) form, simplify that such that we will get the value of the required expression.

Complete step-by-step answer :
Given expression is 3cosec20∘−sec20∘  \sqrt {{\mathbf{3}}} {\mathbf{cosec2}}{{\mathbf{0}}^ \circ } - {\mathbf{sec2}}{{\mathbf{0}}^ \circ }\;
Converting cosec and sec into sin and cos
We get,
=3sin20∘−1cos20∘= \dfrac{{\sqrt {3 }}}{{sin{20}^\circ}} - \dfrac{1}{{cos{20}^\circ }}
Now multiplied and divide the numerator and denominator by 2 we get,
=2(3/2sin20∘−1/2cos20∘)= 2\left( {\dfrac{{\sqrt 3 /2}}{{sin{{20}^ \circ }}} - \dfrac{{1/2}}{{cos{{20}^ \circ }}}} \right)
Now write sin⁡60∘=32\sin 60^\circ = \dfrac{{\sqrt 3 }}{2} and cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}
We get,
⇒2(sin60∘sin20∘−cos60∘cos20∘)\Rightarrow 2\left( {\dfrac{{sin{{60}^\circ }}}{{sin{{20}^\circ }}} - \dfrac{{cos{{60}^\circ }}}{{cos{{20}^\circ }}}} \right)
Taking LCM and solving we get,
⇒2(sin60∘cos20∘−cos60∘sin20∘sin20∘cos20∘)\Rightarrow 2\left( {\dfrac{{sin{{60}^ \circ }cos{{20}^ \circ } - cos{{60}^ \circ }sin{{20}^ \circ }}}{{sin{{20}^ \circ }cos{{20}^ \circ }}}} \right)
Applying sin (A-B) formula we get,
⇒2(sin(60∘−20∘)sin20∘cos20∘)\Rightarrow 2\left( {\dfrac{{sin({{60}^ \circ } - {{20}^ \circ })}}{{sin{{20}^ \circ }cos{{20}^ \circ }}}} \right)
Or
⇒2(sin40∘sin20∘cos20∘)\Rightarrow 2\left( {\dfrac{{sin{{40}^ \circ }}}{{sin{{20}^ \circ }cos{{20}^ \circ }}}} \right)
Now applying sin2A=2sinAcosAsin2A = 2sinAcosA formula
=2×2sin20∘cos20∘sin20∘cos20∘= 2 \times \dfrac{{2sin{{20}^ \circ }cos{{20}^ \circ }}}{{sin{{20}^ \circ }cos{{20}^ \circ }}}
By cancelling the like terms
=4= 4
Hence the value of the expression 3cosec20∘−sec20∘  \sqrt {{\mathbf{3}}} {\mathbf{cosec2}}{{\mathbf{0}}^ \circ } - {\mathbf{sec2}}{{\mathbf{0}}^ \circ }\; is 4
So, the correct answer is “Option C”.

Note : Generally to solve this type of problems what we do is just convert first the given trigonometric expression into sin and cos trigonometric functions then either simplify or use formulas.