Solveeit Logo
Login

Question

Question: The value of \(\dfrac{1}{{\sin {{10}^0}}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^0}}}\) is equals to A. ...

The value of 1sin⁡100−3cos⁡100\dfrac{1}{{\sin {{10}^0}}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^0}}} is equals to
A. 0
B. 1
C. 2
D. 4

Explanation

Solution

In order to evaluate the value of the given trigonometric functions use the trigonometric identities to simplify the equation such as cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b and sin⁡(a+b)=sinacos⁡b+cos⁡asin⁡b\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
1sin⁡100−3cos⁡100\dfrac{1}{{\sin {{10}^0}}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^0}}}
Now take the L.C.M respectively
cos⁡100−3sin⁡100sin⁡100cos⁡100\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 by12\dfrac{1}{2},we get
⇒\Rightarrow 2[12cos⁡100−32sin⁡100]sin⁡100cos⁡100\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. 12=cos⁡600\dfrac{1}{2} = \cos {60^0} and 32=sin⁡600\dfrac{{\sqrt 3 }}{2} = \sin {60^0}
Now substituting the values in the above equation, we get
⇒2[cos⁡600cos⁡100−sin⁡600sin⁡100]sin⁡100cos⁡100\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⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b
Therefore, 2cos⁡70∘sin⁡10∘cos⁡10∘\dfrac{{2\cos {{70}^ \circ }}}{{\sin {{10}^ \circ }\cos {{10}^ \circ }}}
We can write cos⁡700=cos⁡(900−200)=sin⁡200\cos {70^0} = \cos \left( {{{90}^0} - {{20}^0}} \right) = \sin {20^0}
So, now we have,
⇒\Rightarrow 2(sin⁡200sin⁡100cos⁡100)2\left( {\dfrac{{\sin {{20}^0}}}{{\sin {{10}^0}\cos {{10}^0}}}} \right)again, we can write sin⁡20∘=sin⁡(10∘+10∘)\sin {20^ \circ } = \sin \left( {{{10}^ \circ } + {{10}^ \circ }} \right)
Therefore, 2(sin⁡(10∘+10∘)sin⁡10∘cos⁡10∘)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⁡(a+b)=sinacos⁡b+cos⁡asin⁡b\sin \left( {a + b} \right) = sina\cos b + \cos a\sin b
Applying this identity in the above solution we get
2(2sin⁡10∘cos⁡10∘sin⁡10∘cos⁡10∘)2\left( {\dfrac{{2\sin {{10}^ \circ }\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }\cos {{10}^ \circ }}}} \right)
⇒\Rightarrow 2×2=42 \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→Yf:X \to Y examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.