Question

The value of expression $\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$ is equal to
$1){\text{ }}\sqrt 2$
$2){\text{ }}\sqrt 3$
$3){\text{ }}2$
$4){\text{ }}4$
$5){\text{ }}\sqrt 5$

Answer 1

Hint : We have to find the value of the given trigonometric expression$\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$. We solve this using the formula of double angle of sin function . We should also have the knowledge of values for various angles of trigonometric functions . Firstly we find the value of the given trigonometric expression using the formula of sum of two angles of a sine function . Also , the conversion of an angle in terms of another trigonometric function .

Complete step-by-step answer :
Given :
We have to find the value of the expression $\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$
We know , that
$cos{\text{ }}80^\circ {\text{ }} = {\text{ }}cos{\text{ }}\left( {90{\text{ }} - {\text{ }}10} \right)^\circ$
The angle lies in the first quadrant and the value of the cosine function in the first quadrant is positive .
So ,
$cos{\text{ }}80^\circ {\text{ }} = {\text{ }}sin{\text{ }}10^\circ$
Similarly ,
$sin{\text{ }}80^\circ {\text{ }} = {\text{ }}sin{\text{ }}\left( {90{\text{ }} - {\text{ }}10} \right)^\circ$
The angle lies in the first quadrant and the value of the sine function in the first quadrant is positive .
So , $sin{\text{ }}80^\circ = {\text{ }}cos{\text{ }}10^\circ$
Now , the expression becomes
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = {\text{ }}\left( {\dfrac{1}{{sin{\text{ }}10^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{cos{\text{ }}10^\circ }}} \right)
Taking L.C.M. we get
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = {\text{ }}\dfrac{{\left[ {{\text{ }}cos{\text{ }}10^\circ {\text{ }} - {\text{ }}\sqrt 3 {\text{ }} \times {\text{ }}sin{\text{ }}10^\circ {\text{ }}} \right]}}{{\left[ {{\text{ }}sin{\text{ }}10^\circ {\text{ }} \times {\text{ }}cos{\text{ }}10^\circ {\text{ }}} \right]}}
Multiplying numerator and denominator by$4$, we get
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = {\text{ }}\dfrac{{4{\text{ }} \times {\text{ }}\left[ {{\text{ }}cos{\text{ }}10^\circ {\text{ }} - {\text{ }}\sqrt {3 \times } {\text{ }}sin{\text{ }}10^\circ {\text{ }}} \right]}}{{4{\text{ }} \times {\text{ }}\left[ {{\text{ }}sin{\text{ }}10^\circ {\text{ }} \times {\text{ }}cos{\text{ }}10^\circ {\text{ }}} \right]}}
Simplifying , we get
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = {\text{ }}\dfrac{{\left[ {{\text{ }}\left( {\dfrac{1}{2}} \right){\text{ }} \times {\text{ }}cos{\text{ }}10^\circ {\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{2}} \right){\text{ }} \times {\text{ }}sin{\text{ }}10^\circ {\text{ }}} \right]}}{{\left( {\dfrac{1}{4}} \right){\text{ }} \times \left[ {2{\text{ }} \times {\text{ }}sin{\text{ }}10^\circ {\text{ }} \times {\text{ }}cos{\text{ }}10^\circ {\text{ }}} \right]}}
We know that $sin{\text{ }}2x{\text{ }} = {\text{ }}2{\text{ }}sin{\text{ }}x{\text{ }} \times {\text{ }}cos{\text{ }}x$
We also know that
$cos{\text{ }}30^\circ {\text{ }} = {\text{ }}\dfrac{{\sqrt 3 }}{2}$
$sin{\text{ }}30^\circ {\text{ }} = {\text{ }}\dfrac{1}{2}$
Using these values , we get
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = {\text{ }}\dfrac{{\left[ {{\text{ }}sin{\text{ }}30^\circ {\text{ }} \times {\text{ }}cos{\text{ }}10^\circ {\text{ }} - {\text{ }}cos{\text{ }}30^\circ {\text{ }} \times {\text{ }}sin{\text{ }}10^\circ {\text{ }}} \right]}}{{\left[ {{\text{ }}\left( {\dfrac{1}{4}} \right){\text{ }} \times {\text{ }}sin{\text{ }}20^\circ {\text{ }}} \right]}}
Also , the formula of difference of sin function is given as :
$sin\left( {A{\text{ }} - {\text{ }}B} \right){\text{ }} = {\text{ }}sin{\text{ }}A{\text{ }} \times {\text{ }}cos{\text{ }}B{\text{ }} - {\text{ }}sin{\text{ }}B{\text{ }} \times {\text{ }}cos{\text{ }}A$
Using this trigonometric formulas , we get
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = \dfrac{{sin{\text{ }}20^\circ }}{{\left[ {{\text{ }}\left( {\dfrac{1}{4}} \right){\text{ }} \times {\text{ }}sin{\text{ }}20^\circ {\text{ }}} \right]}}
Cancelling the terms , we get
\left( {\dfrac{1}{{cos{\text{ }}80^\circ }}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{\sqrt 3 }}{{sin{\text{ }}80^\circ }}} \right)$$$$ = {\text{ }}4
Thus the value of the expression is $4$ .
Hence , the correct option is$\left( 4 \right)$.
So, the correct answer is “Option 4”.

Note : The various trigonometric formulas we need to remember when by dividing or multiplying the given apply we are able to get a known trigonometric value are given as :
$sin{\text{ }}2x{\text{ }} = {\text{ }}2 \times {\text{ }}sin{\text{ }}x{\text{ }} \times {\text{ }}cos{\text{ }}x$
$cos2x = 2 \times {cos^2}x - 1 = 1 - 2 \times {sin^2}x$
$tan2x = \dfrac{{tan2x}}{{(1 - {tan^2}x)}}$
$sin3x = 3\sin x - 4{sin^3}x$
$cos3x = 4{cos^3}x - 3\cos x$