# Find the value of $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}}$.

Last updated date: 31st Mar 2023

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Answer

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Hint: Put the value of $\cos {45^ \circ }$, $\sec {30^ \circ }$ and ${\text{cosec}}{30^ \circ }$ in the expression and find out its value.

Complete step-by-step answer:

According to the question, we have to calculate the value of $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}}$ .

We know that $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\sec {30^ \circ } = \dfrac{2}{{\sqrt 3 }}$ and ${\text{cosec}}{30^ \circ } = 2$. So, putting all these values in the above expression, we’ll get:

$

\Rightarrow \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{2}{{\sqrt 3 }} + 2}}, \\

\Rightarrow \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{{2 + 2\sqrt 3 }}{{\sqrt 3 }}}}, \\

\Rightarrow \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}} = \dfrac{{\sqrt 3 }}{{\sqrt 2 \left( {2 + 2\sqrt 3 } \right)}} \\

$

Therefore the value of $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}}$ is $\dfrac{{\sqrt 3 }}{{\sqrt 2 \left( {2 + 2\sqrt 3 } \right)}}$.

Note: Since denominator is an irrational number in the above answer, we can also rationalize it to convert it into another form. In rationalization, we multiply the numerator and denominator by the conjugate of the denominator.

Complete step-by-step answer:

According to the question, we have to calculate the value of $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}}$ .

We know that $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\sec {30^ \circ } = \dfrac{2}{{\sqrt 3 }}$ and ${\text{cosec}}{30^ \circ } = 2$. So, putting all these values in the above expression, we’ll get:

$

\Rightarrow \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{2}{{\sqrt 3 }} + 2}}, \\

\Rightarrow \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{{2 + 2\sqrt 3 }}{{\sqrt 3 }}}}, \\

\Rightarrow \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}} = \dfrac{{\sqrt 3 }}{{\sqrt 2 \left( {2 + 2\sqrt 3 } \right)}} \\

$

Therefore the value of $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\text{cosec}}{{30}^ \circ }}}$ is $\dfrac{{\sqrt 3 }}{{\sqrt 2 \left( {2 + 2\sqrt 3 } \right)}}$.

Note: Since denominator is an irrational number in the above answer, we can also rationalize it to convert it into another form. In rationalization, we multiply the numerator and denominator by the conjugate of the denominator.

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