Question

The value of $\dfrac{1}{{\sin {{10}^0}}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^0}}}$ is equals to
A. 0
B. 1
C. 2
D. 4

Answer 1

Hint: In order to evaluate the value of the given trigonometric functions use the trigonometric identities to simplify the equation such as $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$ and $\sin \left( {a + b} \right) = sina\cos b + \cos a\sin b$, using this information will help you to approach the solution of the question.

Complete step by step answer:
According to the question, we have the given equation
$\dfrac{1}{{\sin {{10}^0}}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^0}}}$
Now take the L.C.M respectively
$\dfrac{{\cos {{10}^0} - \sqrt 3 \sin {{10}^0}}}{{\sin {{10}^0}\cos {{10}^0}}}$
Now divide the numerator by 2 and multiply the numerator by$\dfrac{1}{2}$,we get
$ \Rightarrow $$\dfrac{{2\left[ {\dfrac{1}{2}\cos {{10}^0} - \dfrac{{\sqrt 3 }}{2}\sin {{10}^0}} \right]}}{{\sin {{10}^0}\cos {{10}^0}}}$
We know that by the trigonometric identities i.e. $\dfrac{1}{2} = \cos {60^0}$ and $\dfrac{{\sqrt 3 }}{2} = \sin {60^0}$
Now substituting the values in the above equation, we get
$ \Rightarrow \dfrac{{2\left[ {\cos {{60}^0}\cos {{10}^0} - \sin {{60}^0}\sin {{10}^0}} \right]}}{{\sin {{10}^0}\cos {{10}^0}}}$
Since, we know that by the trigonometric identities i.e. $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
Therefore, $\dfrac{{2\cos {{70}^ \circ }}}{{\sin {{10}^ \circ }\cos {{10}^ \circ }}}$
We can write $\cos {70^0} = \cos \left( {{{90}^0} - {{20}^0}} \right) = \sin {20^0}$
So, now we have,
$ \Rightarrow $$2\left( {\dfrac{{\sin {{20}^0}}}{{\sin {{10}^0}\cos {{10}^0}}}} \right)$again, we can write $\sin {20^ \circ } = \sin \left( {{{10}^ \circ } + {{10}^ \circ }} \right)$
Therefore, $2\left( {\dfrac{{\sin \left( {{{10}^ \circ } + {{10}^ \circ }} \right)}}{{\sin {{10}^ \circ }\cos {{10}^ \circ }}}} \right)$
We know that by the trigonometric identity i.e. $\sin \left( {a + b} \right) = sina\cos b + \cos a\sin b$
Applying this identity in the above solution we get
$2\left( {\dfrac{{2\sin {{10}^ \circ }\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }\cos {{10}^ \circ }}}} \right)$
$ \Rightarrow $$2 \times 2 = 4$
So, the answer is 4.
Hence, the option “D” is correct.

Note:
In the above solution we used the trigonometric identities which are the expressions which involve trigonometric functions where the term “function” can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.