What is the value of $\sqrt 3 \cos ec{20^0} - \sec {20^0}$
$
(a){\text{ 4}} \\
(b){\text{ 2}} \\
(c){\text{ 1}} \\
(d){\text{ - 4}} \\
$
Answer
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Hint- In this question we have to find the value of expression $\sqrt 3 \cos ec{20^0} - \sec {20^0}$. This is a basic trigonometric question which can be solved using some basic trigonometric ratios and identities like $\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }}$.
Complete step-by-step answer:
We have to evaluate the expression $\sqrt 3 \cos ec{20^0} - \sec {20^0}$.
Let $x = \sqrt 3 \cos ec{20^0} - \sec {20^0}$……………………. (1)
Now using $\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }}$ in equation (1) we get
$x = \dfrac{{\sqrt 3 }}{{\sin {{20}^0}}} - \dfrac{1}{{\cos {{20}^0}}}$
Taking L.C.M we get
$ \Rightarrow x = \dfrac{{\sqrt 3 \cos {{20}^0} - \sin {{20}^0}}}{{\sin {{20}^0}\cos {{20}^0}}}$
Dividing and multiplying the numerator by 2, above equation can be written as,
\[ \Rightarrow x = \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2}\cos {{20}^0} - \dfrac{1}{2}\sin {{20}^0}} \right)}}{{\sin {{20}^0}\cos {{20}^0}}}\]……………………………… (2)
Using ${\text{sin6}}{{\text{0}}^0} = \dfrac{{\sqrt 3 }}{2}{\text{ and cos6}}{{\text{0}}^0} = \dfrac{1}{2}$in equation (2) we get,
\[ \Rightarrow x = \dfrac{{2\left( {\sin {{60}^0}\cos {{20}^0} - \cos {{60}^0}\sin {{20}^0}} \right)}}{{\sin {{20}^0}\cos {{20}^0}}}\]
Using the trigonometric identity ${\text{sinAcosB - cosAsinB = sin(A - B)}}$ in the numerator part of above expression we get
\[ \Rightarrow x = \dfrac{{2\left( {\sin ({{60}^0} - {{20}^0})} \right)}}{{\sin {{20}^0}\cos {{20}^0}}} = \dfrac{{2\left( {\sin ({{40}^0})} \right)}}{{\sin {{20}^0}\cos {{20}^0}}}\]……………….. (3)
Multiplying and dividing equation (3) with 2 both in numerator and denominator part we get,
$ \Rightarrow x = \dfrac{{4\left( {\sin ({{40}^0})} \right)}}{{2\sin {{20}^0}\cos {{20}^0}}}$
Using the trigonometric identity $2{\text{sinAcosA = sin2A}}$
$ \Rightarrow x = \dfrac{{4\left( {\sin ({{40}^0})} \right)}}{{\sin {{40}^0}}}$
On solving we get
X=4
Thus option (a) is the right answer.
Note – Whenever we face such types of problems the key concept is to have a good grasp over the trigonometric identities, some of them are being mentioned above. This will help you in getting the right track to reach the solution.
Complete step-by-step answer:
We have to evaluate the expression $\sqrt 3 \cos ec{20^0} - \sec {20^0}$.
Let $x = \sqrt 3 \cos ec{20^0} - \sec {20^0}$……………………. (1)
Now using $\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }}$ in equation (1) we get
$x = \dfrac{{\sqrt 3 }}{{\sin {{20}^0}}} - \dfrac{1}{{\cos {{20}^0}}}$
Taking L.C.M we get
$ \Rightarrow x = \dfrac{{\sqrt 3 \cos {{20}^0} - \sin {{20}^0}}}{{\sin {{20}^0}\cos {{20}^0}}}$
Dividing and multiplying the numerator by 2, above equation can be written as,
\[ \Rightarrow x = \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2}\cos {{20}^0} - \dfrac{1}{2}\sin {{20}^0}} \right)}}{{\sin {{20}^0}\cos {{20}^0}}}\]……………………………… (2)
Using ${\text{sin6}}{{\text{0}}^0} = \dfrac{{\sqrt 3 }}{2}{\text{ and cos6}}{{\text{0}}^0} = \dfrac{1}{2}$in equation (2) we get,
\[ \Rightarrow x = \dfrac{{2\left( {\sin {{60}^0}\cos {{20}^0} - \cos {{60}^0}\sin {{20}^0}} \right)}}{{\sin {{20}^0}\cos {{20}^0}}}\]
Using the trigonometric identity ${\text{sinAcosB - cosAsinB = sin(A - B)}}$ in the numerator part of above expression we get
\[ \Rightarrow x = \dfrac{{2\left( {\sin ({{60}^0} - {{20}^0})} \right)}}{{\sin {{20}^0}\cos {{20}^0}}} = \dfrac{{2\left( {\sin ({{40}^0})} \right)}}{{\sin {{20}^0}\cos {{20}^0}}}\]……………….. (3)
Multiplying and dividing equation (3) with 2 both in numerator and denominator part we get,
$ \Rightarrow x = \dfrac{{4\left( {\sin ({{40}^0})} \right)}}{{2\sin {{20}^0}\cos {{20}^0}}}$
Using the trigonometric identity $2{\text{sinAcosA = sin2A}}$
$ \Rightarrow x = \dfrac{{4\left( {\sin ({{40}^0})} \right)}}{{\sin {{40}^0}}}$
On solving we get
X=4
Thus option (a) is the right answer.
Note – Whenever we face such types of problems the key concept is to have a good grasp over the trigonometric identities, some of them are being mentioned above. This will help you in getting the right track to reach the solution.
Last updated date: 20th Sep 2023
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