Question

Evaluate the following:
\[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sec }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+co{{s}^{2}}{{30}^{\circ }}}\]

Answer 1

Hint: We will use the trigonometric ratio table and find out the values of trigonometric ratios as follows:
\[cos{{60}^{\circ }}=\dfrac{1}{2},\sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}},tan{{45}^{\circ }}=1,\sin {{30}^{\circ }}=\dfrac{1}{2},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
Then we will substitute the values of those trigonometric ratios in the equation and simplify to get the answer.
Complete step-by-step answer:
We have been given the expression \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sec }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+co{{s}^{2}}{{30}^{\circ }}}\].
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 \[cos{{60}^{\circ }}=\dfrac{1}{2},\sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}},tan{{45}^{\circ }}=1,\sin {{30}^{\circ }}=\dfrac{1}{2},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\].
So by substituting these values in the equation given in the question, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sec }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+co{{s}^{2}}{{30}^{\circ }}} \\
 & \Rightarrow \dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-{{\left( 1 \right)}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}} \\
\end{align}\]
On simplifying, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{5\times \dfrac{1}{4}+4\times \dfrac{4}{3}-1}{\dfrac{1}{4}+\dfrac{3}{4}} \\
 & \Rightarrow \dfrac{\dfrac{5}{4}+\dfrac{16}{3}-1}{\dfrac{1}{4}+\dfrac{3}{4}} \\
\end{align}\]
On taking LCM of 4 and 3, we get as follows:
\[\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1}{4}+\dfrac{3}{4}}\]
On taking LCM of 4 and 4, we get as follows:
\[\Rightarrow \dfrac{\dfrac{67}{12}}{\dfrac{1+3}{4}}=\dfrac{\dfrac{67}{12}}{\dfrac{4}{4}}=\dfrac{\dfrac{67}{12}}{\dfrac{1}{1}}=\dfrac{67}{12}\]
Therefore, the value of the given expression is equal to \[\dfrac{67}{12}\].

Note: Don’t confuse while substituting the value of trigonometric ratios in the given expression. Also, be careful while calculating and take care of the signs. In this type of question we must remember the trigonometric ratio table so that we can easily substitute any values. In spite of substituting values of \[\sin {{30}^{\circ }}\] and \[\cos {{30}^{\circ }}\], we can also use the trigonometric identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].