Question

Evaluate the following:
\[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+cosec{{30}^{\circ }}}\]

Answer 1

Hint: We will use the trigonometric ratio table and find out the values of trigonometric ratios for the given expression as follows:
\[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},sec{{30}^{\circ }}\dfrac{2}{\sqrt{3}},cosec{{30}^{\circ }}=2\]
Then we will substitute the values of those trigonometric ratios in the equation given in the equation and simplify to get the answer.

Complete step-by-step answer:
In the question, we have been given the expression \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+cosec{{30}^{\circ }}}\].
The angles are all standard angles. So, we can refer to the following table for finding the value of trigonometric functions at standard angles.

Ratio/Angle (\[\theta \])	\[{{0}^{\circ }}\]	\[{{30}^{\circ }}\]	\[{{45}^{\circ }}\]	\[{{60}^{\circ }}\]	\[{{90}^{\circ }}\]
\[\sin \theta \]	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{\sqrt{3}}{2}\]	1
\[\cos \theta \]	1	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{2}\]	0
\[tan\theta \]	0	\[\dfrac{1}{\sqrt{3}}\]	1	\[\sqrt{3}\]	Not defined
\[cosec\theta \]	Not defined	2	\[\sqrt{2}\]	\[\dfrac{2}{\sqrt{3}}\]	1
\[\sec \theta \]	1	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	Not defined
\[\cot \theta \]	Not defined	\[\sqrt{3}\]	1	\[\dfrac{1}{\sqrt{3}}\]	0

Referring to the table above, we can see that \[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},sec{{30}^{\circ }}\dfrac{2}{\sqrt{3}},cosec{{30}^{\circ }}=2\].
So by substituting these values in the expression given in the question, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+cosec{{30}^{\circ }}} \\
 & \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2} \\
 & \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{2\left( \dfrac{1}{\sqrt{3}}+1 \right)} \\
 & \Rightarrow \dfrac{1}{2\sqrt{2}\left( \dfrac{1+\sqrt{3}}{\sqrt{3}} \right)} \\
 & \Rightarrow \dfrac{\sqrt{3}}{2\sqrt{2}\left( 1+\sqrt{3} \right)} \\
\end{align}\]
We have to simplify it further. So, on rationalizing the denominator, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+cosec{{30}^{\circ }}} \\
 & \Rightarrow \dfrac{\sqrt{3}\times \left( \sqrt{3}-1 \right)\sqrt{2}}{2\sqrt{2}\left( 1+\sqrt{3} \right)\times \left( \sqrt{3}-1 \right)\sqrt{2}} \\
 & \Rightarrow \dfrac{\sqrt{3}\times \sqrt{2}\left( \sqrt{3}-1 \right)}{2\times {{\left( \sqrt{2} \right)}^{2}}\left[ {{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} \right]} \\
\end{align}\]
Since we already know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\], we used it to simplify the denominator of the above expression. Now, we can square the terms and we get,
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{2\times 2\left( 3-1 \right)} \\
 & \Rightarrow \dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{4\times 2} \\
 & \Rightarrow \dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{8} \\
\end{align}\]
Therefore, the required value of the expression is equal to \[\dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{8}\].

Note: Don’t get confused in the values of \[sec{{30}^{\circ }}\] and \[cosec{{30}^{\circ }}\]. Sometimes by mistake we may substitute the value of \[sec{{30}^{\circ }}=2\] and \[cosec{{30}^{\circ }}=\dfrac{2}{\sqrt{3}}\] which is wrong. So take care while substituting these values in the expression and substitute \[sec{{30}^{\circ }}=\dfrac{2}{\sqrt{3}}\] and \[cosec{{30}^{\circ }}=2\] properly. For this type of question, we must have to remember all the values of trigonometric ratio for the corresponding angles as shown in the solution.